6 to the (3x + 5) = 1, many don’t know where to start

How to solve an exponential equation using logarithms.
TabletClass Math Academy - TCMathAcademy.com/
Help with Middle and High School Math
Test Prep for High School Math, College Math, Teacher Certification Math and More!
Popular Math Courses:
Math Foundations
tabletclass-academy.teachable...
Math Skills Rebuilder Course:
tabletclass-academy.teachable...
Pre-Algebra
tabletclass-academy.teachable...
Algebra
tabletclass-academy.teachable...
Geometry
tabletclass-academy.teachable...
Algebra 2
tabletclass-academy.teachable...
Pre-Calculus
tabletclass-academy.teachable...

[@quigonkenny]

Start with the 1. 1 = 6⁰, so 6³ˣ⁺⁵ = 6⁰ equates to 3x + 5 = 0, and x = -5/3.

[@user-gr5tx6rd4h]

Yes, I saw this in a couple of seconds and it looks quite ridiculous to use logarithms here, except if the point was just training logarithms.

[@BluesChoker01]

Well, Einstein said it well: "Always describe a problem as simply as possible, but no simpler." Same for programming, don't reinvent the wheel--unless you discover a better wheel."

[@DawnKekana]

@@BluesChoker01do you have any idea on where to start with programming…I tried to instal the python app in my laptop but it declined now I don’t know what to do from here ..

[@girdharilalverma6452]

Yes the simplest and logically correct।

[@boxvism]

That's also how I solved it. But I guess he wants to teach the general method.

[@BleuSquid]

Don't need to use logarithms... formally, at least. You just need to remember the zero exponent rule. Just make the bases the same. By the zero exponent rule, 1 can be rewritten as 6^0. 6^(3x+5) = 6^0 => since the bases are equal, then the exponents must be equal. 3x+5 = 0 => x=-5/3

[@scottaseigel5715]

I said x=-(5/3) within 3 seconds. It’s very straightforward when you know n^0=1 for finite n≠0.

[@paullambert8701]

Yes, I think this a very roundabout way of doing this. He should remember that math tests have time limits.

[@johnwythe1409]

@@paullambert8701 Definitely the quickest way when we know n^0 = 1, but if it is not =1 how do you solve?

[@terry_willis]

@@johnwythe1409 :Yes, this only works if YadaYada =1. If not, then :(((

[@baselinesweb]

Well aren't you just the smartest boy?

[@WitchidWitchid]

@@paullambert8701 If you use logarithms it is also fairly quick. Reason it took so long in the video is because he spent several minutes teaching what logarithms are and how to use them.

[@devonwilson5776]

Greetings. The answer is negative 5/3. The expression 6^(3X+5)=1 can be rewritten as 6^(3X+5)=6^0. Therefore, 3X+5=0 by equating the values of the exponents. Now, finding the value for X is simple. We will transpose the known value 5 across the sign of equality to get 3X=-5, and X=-5/3 after dividing both sides by 3. Lovely.

[@andrewhines1054]

Even less work than my solution - good job!

[@devonwilson5776]

@@andrewhines1054 Greetings. Blessings.

[@ussdfiant]

That’s how I did it.

[@williamniver6063]

Good, correct, and detailed answer, Devon. I am a down, dirty, and done type of guy: (in my head) "Anything raised to the 0 power equals 1, so 3x+5 = 0, x=-5/3, next?) If Math Man John wanted to illustrate how to manipulate exponents and logarithms to solve the more generalized problems as he does in his explanation, he should have used a different problem. By using "1" here, he invites the analysis you and I did, but if it had read "... = 28", the "exponent is 0" trick wouldn't be useful.

[@jim2376]

👍

[@seekerstan]

I'm 72, previously an engineering student, and sea captain. Did the sea captain thing after enginering calculus without a calculator got the best of me. Only to be confronted with the joys of spherical trigonometry that is required to do celestial navigation. I'm just saying you younguns with your GPSs and calculators don't know how easy you've got it. I watch your program to help me from forgeting everything I ever learned.

[@oahuhawaii2141]

But the Earth isn't a perfect sphere. There'll be errors in navigation that needs adjustments to follow the appropriate path to the next port. Otherwise, you could encounter something unexpected and bad.

[@WitchidWitchid]

@@oahuhawaii2141 Which is what they do. Since it's very close the spherical trig gets you quite close to where you need to be and then you can make the adjustments that is part of the navigation process.

[@sporkstar1911]

Fast take - this is a thing-to-zero-power because the result is 1. That means whatever it takes to get 6 to the 0 power. Thus 0=3x+5... ergo -5=3x so x=-3/5

[@AllDogsAreGoodDogs]

Thank you!

[@olivemd]

How I did it too.

[@AquaporinA1]

Your fast take was wrong. You put x=-3/5; the correct answer was x=-5/3. So the moral to the story is to slow down.

[@siamaknormani2733]

Precisely,,

[@dumitrudraghia5289]

ERROR FINAL!

[@masterbaraman9372]

I like the detail you went into here. I think people who want, or need, context will appreciate it, too. Everyone else probably just wants a quick solution or trick. No dig at them, they probably know most of it. I'm just saying not to be discouraged by people who want you to rush. Videos like this help many who need it.

[@DownhillAllTheWay]

The log of a number is the *_power_* to which a *_base_* must be raised to give the number. You can iinvent your own base, but using the example given, log (base 2) of 16 = 4, because 4 is the power to which the base (2) must be raised to give 16.

[@rickcraft816]

So it would seem that many of you are brilliant students who were able to recognize that any number to the zero power is equal to 1 and were able to solve this problem in record time... bravo. What you missed was the brilliance of the teacher to use this problem to be solved using the rules of logarithms, that many may not know or possibly are rusty with, and through his steps have given the student an opportunity to see that these crazy rules actually work and the result actually comes out to a value that we can accept and know to be true. The actual student now has validation of the methods that have been shown result in an answer that they can validate.

[@reeb3687]

generally for equations a^(bx + c) = d, you can take ln of both sides so ln(a^(bx+c)) = ln(d), then use the logarithm power rule to bring bx + c to the outside, yielding, (bx + c)(ln(a)) = ln(d), divide both sides by ln(a), bx + c = ln(a) / ln(d), by logarithm rules ln(a) / ln(d) = log base d of a which i will write as log_d(a), subtract by c to get bx = log_d(a) - c, divide by b for the final answer of x = (log_d(a) - c) / b. if we plug in the numbers, a = 6, b = 3, c = 5, d = 1, x = (log_1(6) - 5) / 3, log base 1 of all positive real numbers (including 6) is 0, so x = -5/3

[@giannaleoci2328]

6^(3x+5)= 6^0 3x+5=0 x=-5/3

[@aryusure1943]

I now know where to start, thanks to your relentless teaching and a lot of practice. Exponential equation? Think Logs! It worked. :)

[@WitchidWitchid]

Anytime you see a more complicated equation or expression in an exponent try using logs to simplify it.

[@aryusure1943]

@@WitchidWitchid Indeed! :)

[@user-tb9fk1jd4q]

set the exponent expression = 0 and solve for x --> 3x + 5 = 0 b/c base^0 = 1 3x + 5 = 0 3x - 5 = -5 x = -5/3

[@jdgower1]

I learned logs back in the day but I didn't even go that long route to the problem, because I also learned that any number to the zeroth power equals 1, so I solved this in my head by just thinking "3x-5=0", and then solving for x - simple math, without using a calculator or log table. I don't care what the base number is.

[@johnplong3644]

How many used Log tables in the back of your Algebra 2 book and use Trig tables??

[@clmkc5393]

Good information and review.

[@jackduncan8561]

Great work, Jon. Can I get into pre calculas with you ?

[@toastmastr9763]

You helped me remember that I know how to do these. Thank you.

[@pennstatefan]

Another method would be to take the log of both sides and one would get (3x + 5)log6 = log1. One would then get 3x + 5 = log1/log6. The next step is 3x = log1/log6 - 5. The solution would be x = (log1/log6 - 5)/3.

[@davidhandyman7571]

I was in my final two years of High School (5th and 6th Forms or Years 11 and 12) here in Australia when scientific calculators became available, but they were equivalent to about a week's pay or more so not many people could afford them. Also, we were not allowed to use any calculator in exams. Calculators were only used to double check our answers and even then, you could not trust calculators to be correct at anything other than basic addition and subtraction.

[@devonwilson5776]

Greetings. I can remember those days very well.

[@davidhandyman7571]

@@devonwilson5776 I also remember using a slide rule and it cost a whole lot less than even a basic calculator.

[@roger7341]

I was a poor graduate student armed only with a slide rule when scientific calculators first became available. I was the only student not using a scientific calculator because I also had a wife and kids to feed. Scientific calculators should never be banned from any exam when they are not banned from the job. Instead, the student should be required to justify every step taken to arrive at any answer obtained from a scientific calculator. For example, suppose my boss handed me this new equation he just discovered, (2.3)^(2x)-4*(2.3)^x+4=0, and ordered me to have a value for x by tomorrow. And by the way, scientific calculators are not allowed. I know several ways to solve this problem, but I can't think of a useful way to get a value for x without a scientific calculator. For example, substitute y=(2.3)^x into the given equation to get y^2-4y+4=0, which has roots y=[4±√(16-16)]/2=2, so (2.3)^x=2. Of course I could then take the natural log of this equation to get x*ln(2.3)=ln(2) and solve for x=ln(2)/ln(2.3), but without a scientific calculator I couldn't provide a numerical value for x without consulting my old "CRC Standard Mathematical Tables and Formulae" book.

[@oahuhawaii2141]

If you can't trust those calculators, then they're useless. What you couldn't trust is the user, if he/she didn't understand how the device operates and its limitations. I think the real problems are that most early ones had no support for extended precision, floating point (mantissa & exponent), precedence rules, parentheses, and memory store/recall. They also lacked many common functions, such as roots, powers, exponentiation, logarithms, circular & hyperbolic functions along with their inverses. That means you still need to make sure you don't exceed the range of the calculator, use look-up tables for the common functions, be wary of precision errors introduced, and have scratch paper on hand to record intermediate results that will be hand-entered later on. That's where errors get introduced into the calculations. My sister got a TI SR10. It has built-in [x²], [√x], and [1/x] functions, but no parentheses or memory store/recall. It can display the result in mantissa with exponent format. Neatest thing since sliced bread for engineers and scientists.

[@adrianm.2043]

Hi. Great to see this. I must have done it at school but I don't remember it. As for pocket calculators they were not available until maybe 1970 when Sinclair started selling a pocket calculator at an affordable price, (in the UK) this calculator did only basic arithmetic, but very soon Texas instruments brought out a significant scientific programable calculator, a little too large to fit a shirt pocket it had most of the features you might expect today. I bought one to use at work in about 1972/3, it meant calculations that previously needed seven figure logarithm tables and took hours now took minutes. At the time I was working as a design draftsman and still an apprentice. The calculator literally cost a weeks pay. That calculator lasted until the early 1990's when I had the chance to do an engineering degree, I bought a new one then, it cost about the same in money as the previous one. I see today a similar Texas Instruments calculator costs much the same, which in real terms is many times cheaper.

[@oahuhawaii2141]

A friend got a Sinclair calculator. It was fun to play with, but we kept draining the batteries. He also got the computer, too.

[@oahuhawaii2141]

My sister got a TI SR10, and later my brother got the HP 34C. I got a TI 35, TI 57, and HP 41C. Another friend got a TI 58. Another student got the TI 59. They were fun to use.

[@fjp3305]

Calculators came in the mid 70s, and they were very expensive. And you are a very good teacher. Thanks.

[@malenor4148]

I took apart my 1970s Texas Instruments calculator and activated the square root button.

[@oahuhawaii2141]

My sister got a TI SR10. It has built-in [x²], [√x], and [1/x] functions, but no parentheses or memory store/recall. It can display the result in mantissa with exponent format. Neatest thing since sliced bread for engineers and scientists.

[@jerry2357]

6^(3x+5)=1 so taking logs of both sides and dealing with the exponent we get (3x+5)*log(6)=log(1), and thus 3x+5=log(1)/log(6), but log(1)=0 (for logarithms of any base) so 3x+5=0, so x=-5/3.

[@kennethwright870]

For that equation to be true, 3x+5 must equal 0, so x=-5/3

[@bigc1903]

I am so happy I never had you to teach me math.

[@OR161NYT]

you funny man, i like u

[@BillintonAllen]

Hi John you are an excellent teacher. I am having a lot stuff cleared up

[@ronnieq9706]

I got the answer by accident. I learn so much from your videos. Thank you!

[@victorajayi2845]

What application do you use for your presentation. What application do you write abd draw on please?

[@STRIKER.OP44]

Sir you are underrated you deserve more views and subs

[@pauldalnoky6055]

Views is what count$

[@BruceKarger]

No need for log. Any number to the zero power is always 1. So, 3x+5=0. 3x is -5. Divide bot side by 3. X equals -5

[@clmkc5393]

Good info!

[@pmw3839]

Yes, log tables and all the other tables were extremely tedious to work with. Calculators are a blessing. As is KZbin and computers and smart phones. We forget how lucky we are these days.

[@paullambert8701]

I still like slide rules.

[@pmw3839]

I like your quick crash courses.

[@warrenstanford7240]

Went to school from 1972 through to college leaving in 1985 at eighteen and parents bought me a Casio scientific calculator in 1981 for my Maths O levels and Physics.

[@user-cx5dm2ch4v]

On Indices any number raised to zero gives 1 therefore the power on the right is zero meaning equating 3x+5=0 solving gives X=-5/3 .The explanation may confuse learners.

[@psirotta]

Simpler: If (3x + 5)log 6 = log 1 = 0, then since log 6 not= 0, 3x + 5 = 0, and x ≈ -5/3.

[@v8pilot]

Take logs: (3x +5) × log 6 = log 1 (note that log 1 = 0) therefore (3x + 5) × log 6 = 0 divide each side by log 6: 3x + 5 = 0 so x = -5/3 = - 1 ⅔

[@Silentwarzone]

the inverse of exponent is logarithm, Why do i keep thinking is square rooting? Whenever i rearange a formula with exponent, i do sqr rooting on both side as inverse or? Sir can you do a vid on the difference between the two please.

[@ndailorw5079]

You’re right, when and where it applies. For example, x^3 = 27, which gives x = 3 after taking the cube root on both sides of that equation. Or, the same thing for x^2 = 100, say, which, taking the square root on both sides gives x = +/-10, apart from dimensions of time, length, etc., since there’s no such thing as negative length and time, etc. in which cases only the positive answer and not the negative answer is possible. Or, and probably more to your question and example but the very same thing nonetheless, and simply the same thing the other way around but using the inverse of roots we could have the square root of x = 10 where we’d then square both sides of the equation (the inverse of taking the square root on both sides) to get x = 100! Or we could have the cubed root of x = 3, where we’d simply raise the expressions on both sides of the equation to the third power (cube both sides of the equation, sort of speak) to get x = 27. Well, here, the same thing’s going on since logarithms and exponentials are inverses of each other, like exponents and roots are, one to the other, just as addition is to subtraction and division is to multiplication, and vice versa, etc. That is, we can use the inverse of a mathematical operation that’s in the form of an equation to solve that equation, or to check for the correctness of it. In other words, we check our answers to our addition for their correctness by its inverse operation of subtraction, and vice versa. Same with multiplication and division with respect to each other, etc.

[@jerryclasby9628]

6^(3x+5) = 1 = 6^⁰ Therefore 3x+5 = 0. Exponent values are equal 3x= -5 X = - 5/3 QED

[@eathan_ozawa]

Is someone able to build a solution in Excel/Google? I'm having a hard time solving for a complex exponent in Google/Excel. This is based on a financial formula. A=P*((1+r/n)^(n*t))+x Solving this for (t) (A-x)/P =(1+r/n)^(n*t) At this point, I think I need to use a log function to get the exponent out, but if I capture (t) in "log(1+r/n,n*t)", I'm really not sure what to do to get (t) out of the function. Please help!

[@russelllomando8460]

wow, i actually deduced that the exponent had to be 0 for the equation to = 1. dang, sometines i amaze myself. all in fun. thanks for a great lesson, & actual reassurance. i went to HS in the mid-late 60s. if you got caught with a calc, YOU FAILED, sometimes the test or even the semister. DONE !!!

[@user-ri6rn7ti5h]

All good thing comes in three's

[@devonwilson5776]

Greetings. I remember those days, for me 70's, in primary school, absolutely no calculator allowed. The good old days.

[@ndailorw5079]

That’s because calculators are for students who understand and therefore know how to do the mathematical calculations without using a calculator. What if the calculator broke down and we didn’t know how to arrive at the answer way the instructor is doing here? It’s easy to hit 5 and then log10 on the calculator. But what if the calculator pegs out and break right at that moment and we didn’t know how to do the problem by hand because we didn’t know and understand how to do the problem by head and hand? Calculators are excellent for doing tests and even during class, but only after a subject has been thoroughly understood and learned by the head and done by the hand repetitiously until committed to both memory and understanding. Does no good to have students hit and peck 2 + 2 on the calculator when they don’t understand and know how to do addition through understanding and repetitive practice. What happens if the instructor gives a set of problems to be done within a reasonable time frame but says, “absolutely no calculators!” Then what! Calculators are for those who understand and know the work, they’re not for the uninitiated! The uninitiated will never understand and learn the subject matter properly and thoroughly by simply punching in log100 with a calculator… they need to first understand and learn why the calculator gives the answer 2! For example, why would I add 0.3456889731237 + 35.903578134556 = x by head and hand instead of using a calculator when I already understand and know how to add decimals by head and hand? It would consume way to much time and be completely insane for the teacher to have me to write that problem out by head and hand instead of using a calculator when the teacher knows that I’ve proven and therefore understand and know how to do the problem by head and hand! But for the uninitiated student they must necessarily begin by first understanding and knowing and doing the problem of adding decimals before they move on to calculators. Calculators are horrible for students who don’t first understand know how to do math by head and hand through practice, practice, practice! So… what do I do with 30.4567893400586 - 7.8967855422106 if I didn’t know how to subtract decimals by head and hand but knew how to peck out the answer on a calculator but all the calculators in the world were broke….?

[@abeshdas6130]

Its a easy solution Take 1 = 6^0 bcoz anything power to 0 is 1 Therefore 3x+5 =0 X= -5/3 No need for this long process

[@BluesChoker01]

Yep, when the bases are identical, they can be eliminated from exponential equations. Of course, this is a *nice* example because "1" can be replaced by 6^0. The change of base formula is unnecessary. 😊 To generalize, if equation were: 6^(3x+4)=36 then 6^(3x+4)=6^6 3x+4=6 3x=2 x=2/3 Chk: 6^(3*(2/3)+4 )=6^6 6^(2+4)=6^6 6^6=6^6 36=36 This identity holds for both real, complex and trigonometric exponents.

[@dianegesik7456]

Thanks for providing the explanation. It is helpful. I am arriving at a different answer than yours however. Could you take a look to see if you agree? In the third line of your calculation you are showing 6^(3x+4)=6^6. I am asking if this should instead be 6^(3x+4)=6^2. Thus 3x+4=2, 3x=-2, and x=-2/3.

[@roger7341]

I'll solve this equation in three different ways, and one should know all three of them. (1) 6^(3x+5)=6^(3x)*6^5. Divide both sides by 6^5: 6^(3x)=6^(-5). Equate powers of 6: 3x=-5 so x=-5/3 (2) Any number raised to the zeroth power is 1, so 6^0=6^(3x+5)=1 yields 3x+5=0 and x=-5/3 (3) The natural log of 1 is 0, so take the natural log of both sides: (3x+5)ln(6)=ln(1)=0, so (3x+5)=0/ln(6)=0 yields 3x+5=0 and x=-5/3

[@Astrobrant2]

Maybe you should have had this equation equal something other than 1, since knowing that anything to the zero power =1. So if 3x+5 is zero, then it's really easy and doesn't require logs. Anyway, thanks for the instruction on logs. They always kinda perplexed me, and you're helping me to understand.

[@vipulakularathne3664]

Good lesson

[@RoyPierce-fb8mt]

Don't forget to mention the slide rule!

[@BluesChoker01]

We used slide rules when I was a junior, but we could use calculators as seniors. The school bought them and we could borrow one. The first Calc I used was the HP which used polish infix notation. It did everything but graph. Switched to TI as undergrad. Returned 20 years later for advanced degree, and we used a nice--TI graphing calculator. But for advanced Calc, discrete math, etc--we used laptops and Maple and Mathematica (now Wolfram). Once mastering these tools, you just couldn't miss homework problems. The best thing about these tools, was the ability to try all sorts of approaches. If an approach failed, you learned something. Playing around and thinking really expands your depth of knowledge. One thing to remember is there are no timed tests in the workforce-- only deliverable dates. Sure, some things needed quickly reviewing, but some problems took months, so they were always in the background, to be attacked as insight lit up connected neurons :-) Why is it that most "aha" moments occur in the shower, working out or in dreams?? 😮 I think the subconscious is handling a lot of work so the conscious mind can function on the immediate tasks at hand. Man, I got pretty metaphysical here.

[@carolbattle3389]

Back in the good old days you used a slide rule. I used tables and a slide rule in 1955.

[@oahuhawaii2141]

Yes, but in his 2 problems you can do them easily, right? 6^(3*x+5) = 1 Take log base 6 of both sides: 3*x+5 = 0 x = -5/3 ≈ -1.666667 4^y = 10 y*log(4) = log(10) y = 1/2/log(2) We remember log(2) ≈ 0.30103, so y ≈ 1/0.60206 ≈ 1.660964 .

[@marydefang6993]

Any number raised to the zero power equals 1. So if 6 raised to the power 3× +5=1. It implies that 3× +5= 0 and when you solve this ,it gives × = - 5

[@nickcellino1503]

No. It's -5/3. You forgot to divide 3x by 3.

[@jim2376]

Start with the exponent. It has to yield 0 since 6^0 = 1. So 3x + 5 = 0. x = -5/3

[@nycstreetpoet]

I’m not sure how to solve it, but I know the exponent needs to be 0 fir this to make sense. I divided by 3 and got -5/3, but this was an expression, not an equation, and in an expression I can’t extricate the variable from it, as I understand it.

[@keithschipiour4684]

Yes many won’t no where to start because they preoccupied by the equation of how to make there paycheck equal to the cost of living.

[@kumardasi]

6 to the power of zero equals 1. Hence, 3x+5=0; this yields x=-5/3

[@norcalovenworks]

6 to the zeroth power is equal to one. Easy to solve now, right?

[@Tzuriah]

Yeah, the ‘70s……….we had to actually DO this lol! Ah, the tables in the back of the books. We were allowed to use the NEW scientific calculators in my senior year (1978). They were exorbitantly priced (over $100)! The minimum wage was $2.65 then. It would be like paying over $275 today. My dad was unhappy! I had a friend who used his trusty slide rule. Ah, those were the days!

[@davidhandyman7571]

I am nearly 70 years old. There are "gaps" in my math. These are because of poor teaching when I was in school. Just in case you think I am/was the problem, I was always at or near the top of my class in Math and achieved being in the top 10% of the State in my Higher School Certificate. Having someone to actually teach you correctly is essential. By the way, I correctly got the answer in about 15 seconds without using logs.

[@rremnar]

I can vouch for this. When I went to Lake Stevens HS, the geometry teacher was a total bore. I didn't learn much and I had a hard time paying attention, so I would up with a D grade. When I transferred over to Mariner HS, I got into a trigonometry class. The teacher there was good, and I got a B. The teacher does make a huge difference, but the door swings both ways.

[@WitchidWitchid]

At first it looks like a hard and tricky problem due to the x variable in the exponent. But it is really a very easy question.

[@FormlessFlesh]

Answer: -5/3. Take the natural log of both sides, bring the exponent down on the left side. (3x + 5) ln (6) = ln (1) ln(1) is equal to zero: (3x + 5) ln (6) = 0 Divide both sides by ln (6): 3x + 5 = 0 Subtract both sides by 5: 3x = -5 Divide both sides by 3: x = -5/3

[@FormlessFlesh]

You can also take the log instead of ln. Either way, you get the same answer

[@user-oo2qs7ev7p]

A^0=1 when bases are same powers are equated so 3x+5= 0 , so x=-5/3

[@Dan_Neely]

Log tables were only for when you needed more precision than your slide rule could give.

[@dahcargo]

I got -1.66 why is that wrong? Should I have left my answer in a fraction? I know it means nothing, but, this is fun. I've had much higher calculus than any of this, but I apparently have a lot to learn?

[@haroldkingdvgalatierra2552]

It's easy, exponential and logarithmic

[@shadrana1]

6^(3x+5)=1=6^0 compare indices, 3x+5=0 x= -(5/3). It took me about 3 secs.

[@daleleppert6914]

I typed into my calculator 'log10÷log4 and the answer was 1.22 weekday am i doing wrong?

[@vespa2860]

Using log to base 10 ( the normal log button): log 10 = 1 log 4 = 0.6 approx gives 1.66 when divided Using natural log (the ln button): log 10 = 2.3 approx log 4 = 1.386 approx again gives 1.66 I've no idea how you arrived at 1.22 (even tried mixing the two log functions). edit - I used your 1.22 and swapped around to find either: log 10/log 6.6 = 1.22 or log 5.426/log 4 = 1.22 Doesn't help much! On the Windows calculator (scientific mode) Simply input 10 press log - then divide by 4 press log i.e. (10 log)/(4 log) Not sure what calculator you are using.

[@ndailorw5079]

…may depend on the hierarchical structure (program) built into your calculator. On mine I hit 10 (the 1 and 0 buttons), then hit the log 10 button, then hit the divide button, then hit the 4 button, then hit the log10 button, then hit the = button and the calculator gives the approximation 1.66……….. how you enter those terms in the proper sequence and correctly on a calculator depends on the calculator’s hierarchy structure.

[@okonol]

This can be solved using derivatives, right?

[@user-bo5yt1ub1o]

i entered log10/log4 and my calculator showed 1.2203,,, as answer not 1.66. whats up? maybe i entered wrong syntax?

[@ndailorw5079]

Calculators aren’t all programmed the same. Some accept entries one way while others do it differently. Find out what type of hierarchical program your calculator has, or look at its usage instructions and you’ll get the correct answer.

[@jamesharmon4994]

Step #1, convert 1 into 6^0, then with the same base, exponents must be equal.

[@Mike-lx9qn]

3x+5 = 0, 3x=-5, x=-5/3.

[@Mike-lx9qn]

Especially easy when it's equal to one & only done so when it's equal to one. You can also do logs on both sides & doing a log of 1 is equal to zero & a log base 6 of 6^z gets you z = 0 |z=3x+5 & u get the same thing

[@jamesharmon4994]

@Mike-lx9qn Yes, but it seems unnecessarily complex and does not teach the principle that when bases are equal, then the exponents must be equal.

[@thomasmaughan4798]

@@jamesharmon4994 "bases are equal, then the exponents must be equal." However, 5^0 = 1, 6^0 also = 1, so it is not clear that one must choose 6^0. Instead it is probably sufficient to know that any number to the 0 power is 1, thus the exponent must be equal to zero and it isn't really necessary to put the same base on both sides of the equal.

[@jamesharmon4994]

@thomasmaughan4798 True, but if the instructor requires "Show your work", it is.

[@sathyarajan100]

6^3x+5= 1 6^3x/6^5=1 6^3x =6^-5 base 6 is the same on both sides 3x = -5 X =(-5/3) For checking this value in the above equation 6^3(-5/3) +6= 1 6^(-5 ) +5= 1 6^0= 1 1 =1

[@DawnKekana]

What happend to the 1 when you moved to step 3?

[@pauldalnoky6055]

Natural logs of both sides, John?

[@richardl6751]

He is using common logs, base 10.

[@N269]

Makes no difference.

[@oahuhawaii2141]

Take log base 6 of both sides to get 3*x+5 = 0. Then, we see x = -5/3 .

[@oahuhawaii2141]

If you want to go for the infinite solution route with complex numbers, then you need to consider the polar form for 1, which is 1*e^(i*2*π*k) for all integer k: 6^(3*x+5) = 1 = 1*e^(i*2*π*k) (3*x+5)*ln(6) = 0 + i*2*π*k x = -5/3 + i*k*(π*2/3/ln(6)), for all integer k Note that for k = 0, x = -5/3 .

[@Paul47088]

The full solution should be -3/5 + 2*n*pi*i , where is is square root of (-1). n is any integer

[@oahuhawaii2141]

Nope. You are way off.

[@oahuhawaii2141]

If you want to go for the infinite solution route with complex numbers, then you need to consider the polar form for 1, which is 1*e^(i*2*π*k) for all integer k: 6^(3*x+5) = 1 = 1*e^(i*2*π*k) (3*x+5)*ln(6) = 0 + i*2*π*k x = -5/3 + i*k*(π*2/3/ln(6)), for all integer k Note that for k = 0, x = -5/3 .

[@saadykohanim7489]

Most simple way is to make the power of 6 zero, so because 3x +5 to be zero needs 3x to be equal to -5 so x should be -5/3

[@jw3946]

A much easier way. 6 to the zero power is 1. As such 3x + 5 = 1. Took me less than 3 seconds to solve.

[@markmurto]

Any number to 0 power is 1 3x+5=0 done.

[@hannukoistinen5329]

Just start taking logs from both sides, which eliminates the 6, and solve the equation.

[@user-ri6rn7ti5h]

4^=10= 3 .2 (x+2x-3)

[@martinstabb7059]

Surely 6^0=1, so with a common base of 6 we have 3^x+5=0, ie x= -5/3

[@pennstatefan]

could you use x = -(5/3) to solve the problem?🤔

[@luisguzman6426]

If we assume that any number that has 0 as an exponent is equal to 1, then 3x+ 5 = 0

[@SB9854]

if u use the a^0=1 then 3x+5=0--> 3x=-5-->x=-5/3 isn't it easier?

[@harrymatabal8448]

At the beginning would be excellent

[@DuyenNguyen-cx7do]

Any number not zero, to the power of zero equal 1. Thus 3x+5 = 0, or x = -5/3.

[@jamesholden4571]

For any y, y^0 = 1 Therefore, 3x + 5 = 0 3x = -5 x = -5/3 Where is the challenge? I did this in my head.

[@nelsonfilho6779]

Parece fácil : 1=6^0 assim 3X+5=0 X=-5/3 e pronto !!! 🤷🇧🇷

[@mossleopeng1928]

1 = x^0 , therefore 6^(3x+5% = 6^0, therefore 3x+5=0 x=5/3

[@stelun56]

JEEZ NO WONDER STUDENTS HATE MATH 25 MINUTES FOR THIS

[@Pgan803]

It is Not the Answer that you are learning here. Its the Method so can apply to every others. In Maths Methods solves.

[@tevman69]

So, how is this applied in the real world?

[@user-tl9bq7gd9v]

But you didn't show why anyting raised to zero is one; i.e. why x^0 =1 - I will show it in two ways, the first with numbers to help get the feeling for it, and the other with a in order to "make it mathematically correct as a proof" 16=2^4, 8=2^3, 4=2^2, 2=2^1, 1=2^0, 1/2=2^-1, 1/4=2^-2, 1/8=2^-3, 1/16=2^-4 Beginning with 16, every time you divide by two, the result is that one is subtracted from the exponent, and so your exponent starts at 4 and is reduced to 3, 2, 1, 0, -1, -2 -3, -4 while the number is halved down to 1 and then to 1.2 to 1/4 etc, and the 1 occurs when the exponent is 0. axaxaxa=a^4, axaxa=a^3. axa=a^2, a=a^1, a/a=a^0, 1/a=a^-1, 1/(axa)=a^-2, 1/(axaxa)=a^-3, 1/(axaxaxa)=a^-4. Take a4/a =a3 (axaxaxa)/a =(axaxa). Similarly (a^3)/a=a^2. Similarly (a^2)/a=a^1. Similarly (a^1)/a=a^0. Similarly (a^-1)/a=a^-2. The crucial point is (a^1)/a=a^0. i.e. a^1 = a, and a/a=1 while a^1/a^1 = a^1 x a^-1 = a^(1-1) = a^0. So we have a/a = a^0 and a/a=1

[@billh5923]

I started with x = -5/3 Done

[@juancarlosnadermora716]

There’s another way and is elevate both terms of the equation to 0 And so forth and so on.

[@robert8552]

Multiply both sides with 6^-5 => 6^3x = 6^-5 => 3x=-5 .............

[@gapecanbrown]

Start with bringing the exponent down

[@user-bn7ud5wp2t]

How about this 6^3x+5=6^0, then 3x+5=0 and x=-5/3... Fast and easy

[@jjjilani9634]

Change 1 with 6^0

[@m.fferguson9787]

One of the easier ones. Everything raised to zero gives one, thus the parentheses must give zero and thus x=-5/3. If you take log10, either log6 or 3x+5 must be zero before that to the right of the equals sign can become zero

[@leekimmelman1069]

Why does it take 25 minutes to explain that any thin raised to the “0” power is equal to 1?

[@BurninVinyl]

I would just set 3x+5 as t and solve t first and then substitute it.

[@robhaver8704]

whenever an answer to a question is approx. i sign off. that's not an answer, that's a guess.

[@onetime1239]

I knew that the only way that 6 to an exponent would equal 1 is that the exponent would have to be zero, so i solved 3x+5=0 which gave me -5/3.

[@MrBobbybrown7]

any number raised to the power of 0 is 1, therefore 3x +5=0 and QED x=-5/3

[@oahuhawaii2141]

Some mathematicians will squawk about 0^0 .