(2 to the x - 1 power) divided by (2 to the 3 - 4x power) = 16, many don’t know where to start

How to solve an exponential equation with no calculator.
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[@sergiussouth4754]

I just found a 1936 edition of; Mathematical Tables from Handbook of Chemistry & Physics. It's 307 very thin, pages of pure mathmatical gold. Wow.

[@robertellis3915]

In the 80s in high school with a lot of learning disabilities in California i was never tought math like this

[@russelllomando8460]

Great lesson. My HS algebra was in 1965.

[@ephraimgarrett4727]

1966 for me. In all the years since then, I never once needed to know how to do this. 😄

[@b213videoz]

It's ok, algebra hasn't changed since then 😊

[@b213videoz]

​@@ephraimgarrett4727Nice to be an American living a typical American life 😂

[@devonwilson5776]

Greetings. The answer is 8/5. Without the use of a calculator, we will rewrite the expression as 2^x-1 -(3-4X) =2^4. Now we will equate the indices to get X- 1 -3+4X= 4. Moving on, we will get (X +4X) = 4 +3 +1 after adding all the X's and transposing the known values to the opposite side of the X's resulting in 5X=8. Therefore, the value for X that satisfies the equation is 8/5.

[@Jasper7182009]

A nice, straightforward math problem!

[@johnny3433]

Yes easy Problem

[@martinfiedler4317]

Good bit-sized practice with excellent explanation. Thanks!

[@afre3398]

A logarithm is the power to which a number must be raised in order to get some other number. And the logarithm to the base number is always 1. The base number can be any number. But 10 (log) and natural logarithm(ln) base e ≈ 2.718 is the most common. So what TabletClass Math did here by experience. Was to take the log(base 2) on both sides. And ended up with log(base 2)(2)x(5X-4)=log(base 2)(2)x4. Log(base 2)(2) is 1. So it goes away. And you end up with 5X-4=4. Also if you have an equation like A(5X-4)=A(X-3) the A will cancel out on both sides. So if you had chosen to take out the big hammer and done log(2)x(5X-4)= log(2)x(4). The log(2) on both sides would have gone away and no need for calculator at all

[@garyalabama]

Amazing how difficult I thought this was in high school but so easy now.

[@QThePrimary]

When you spoke that textbooks can not translate in words all that goes wrong is so correct. C.P. Snow wrote an article in 1948 The Two Polars. My Higher Self so happened to show me the article in college. Same year, I have a math textbook and first day of class the teacher RANDOMLY said 'this is a textbook you cannot read'. I didn't understand what he said. Life happens. I'm a returning student years later. First day of class the math professor says, 'this is a textbook you can read'. I looked at it as a Blessing, which it was. The Two Polars is a LONG article on how there was a seminar in Edinburgh. And he realiZed the humanities didn't know the sciences and vice versa and how there is a gap in regards to teaching in school system {basically}. Quantitative vs Qualitative.

[@danieldennis9831]

x=8/5 First I used a wrong rule (I did a^m•a^n=a^(mn)) and got x=13/17 Then I saw that my answer didn't match the video and I tried the a^m/a^n=a^(m-n) and got x=6/5 Then I saw that I was concentrating so much on making certain the minus was distributed to the -4x that I accidentally also re=switched the 3. So I redid it again. 2^(x-1)/2^(3-4x)=2^4 2^[(x-1)-(3-4x)]=2^4 (x-1)-(3-4x)=4 x-1-3+4x=4 5x-4=4 5x=8 x=8/5. That was the right answer. And then I remembered the first rule correctly (a^m•a^n=a^(m+n) and redid it the first way I thought of: 2^(x-1)/2^(3-4x)=2^4 2^(x-1)=2^4•2^(3-4x) 2^(x-1)=2^(4+3-4x) x-1=4+3-4x 5x=8 x=8/5. And that is the right answer as well. My mistakes were: Rushing, mixing two rules (a^m•a^n=a^(m+n) and (a^m)^n=a^(mn)) and not being mentally diligent in the steps.

[@WombatMan64]

Hmmm. I get your point about not having a calculator so logs aren't useful here.... except they are if you use log base 2 in a problem containing powers of 2. It turns the 16 into a 4 because log base 2 of 16 is 4 (easy enough to do in your head). 2^(5x-4) becomes log(2) 2^(5x-4), move the exponent to get (5x-4) x log base 2 of 2... log base n of n is 1, so you're left with 5x-4=4. But at the end of the day your way was a touch quicker here.

[@mathewjohn8126]

Thanks 🙏

[@michaelmappin4425]

Two more tools in my math toolbox. Thank you

[@TheDerlick]

I actually first resolved the denominator by multiplying both sides by 2^(3 - 4x). Same answer, slightly different way. 2^(x - 1) = 2^4 • 2^(3 - 4x) 2^(x-1) = 2^(4 + 3 - 4x) 2^(x - 1) = 2^(7 - 4x) x - 1 = 7 - 4x 5x = 8 x = 8/5

[@lizardfirefighter110]

This is all fine and dandy, but how would a word problem read for this set up?

[@frankjoiner7661]

For example, if if “something10” is divided by “something4,” the answer would be “something6.” What could “something” be.

[@aryusure1943]

By way of review I came back to this problem and I have to say that I made the "common mistake" you talked about again. But besides that, I was on my way. Good enough for me but I need to distribute better next time. ;) Because I had a minus 4 instead of a positive 4 in my exponent I came up with this answer: - 8/3. Close but I see what I did wrong.

[@raya.pawley3563]

Thank you

[@Brovider]

Keep going please example after example you teach me better than my teacher

[@Brovider]

Please continue i will lose my scholarship post all the equations existing

[@vee8217]

First step Change 16 to base 2 to the power of 4. 2^4= 16 Now cross multiply 2^ 3-4x multiply 2^4 Delete base 2 3-4x+4= x-1( from the top equation) Then -4x-x= -1-3-4 -5x= -8 X=8/5 I know it's a bit too long.🙉

[@Sepia36912]

Hello John.

[@robertellis3915]

55 guy here in klamath Falls Oregon

[@user-te8gy9cw6h]

I was good at math in high school. I now somewhat regret not utilitzing that strength when I got into College. I advise others not to make that same mistake.

[@harrymatabal8448]

Will take your advice. Will not go to college 😂

[@luisPcordeiro]

I tried to apply the logarithmic rules first to make 2^(x-1) become (x-1) . log(2) and the same to the denominator (3-4x) . log(2), thus cross canceling the log(2). Rule "The logarithm of a number raised to a power is the power times the logarithm of the number." However, I did not get the correct result. Instead, I got x = 49/65 I wonder why; I should be able to apply the logarithmic rules immediately. Right? in summary 2^(x-1) / 2^(3-4x) cannot become this below (?!) following the rule? ( (x -1) . log(2) ) / ( (3 -4x) . log(2) ) The idea is to be able to cross-cancel the log(2) since it's a factor in a fraction? Thinking about this...

[@luisPcordeiro]

I got why. I cannot make that transformation, because the expressions are not equivalent. It's a bit hard to explain, and I do not feel capable of doing it. Let's just say that 2^3 is not the same as "". It needs to apply to both sides example 2^3 = 8 so 3 . log(2) = log (8) which is a subtle difference :)

[@harrymatabal8448]

​@@luisPcordeirowhy would you want to use log in the first place

[@katherinei7827]

Where did the minus sign come from: - (3 -4x)

[@andtrrrot]

When dividing exponents with the same base, you subtract the exponents. Thus, x-1- (3-4x) . Distributing the negative, you get -3+4x.

[@frankjoiner7661]

What I’m trying to imagine is circumstance where this formula would be applicable. What could x be? Whatever it is, you already some crazy manipulations equal 16. But 16 what.

[@arthurburke1044]

Where’s the road Map 🗺????

[@benquinneyiii7941]

Unless they don’t like improper fractions

[@harrymatabal8448]

2^(5x-4) =2^4....x=8/5

[@aryusure1943]

The day I will stop struggling with negative signs in front of parentheses will be a happy day. ;)

[@kencope1984]

Many might not know - but many more do not care

[@oldschoollew]

Again id had the approach correct and lots of problem simplifying.....

[@johnnolen8338]

Ans. x = 8/5.

[@orangebetsy]

I don’t understand 8/5 as an exponent…how do you raise 2 the 1.6 power? I know it’s ^8/5 - 1 and ^3 - 4(8/5) but I’m not very advanced in math so the question remains

[@ndailorw5079]

@orangebetsy3404 First, just to be mathematically clear, it’s 2^[(8/5) - 1], and, 2^[3 - 4(8/5)]. And we’re not raising 2 to the 8/5 power! 8/5 is simply the answer for x. To check our answer for correctness by substituting our answer, 8/5, back into our equation what we are really doing, and must do, is raising 2 to the power of what results from the calculation done within the parentheses. We’re not raising 2 to the 8/5 power! Careful! But more to your fundamental question, as you know, decimals and percentages are fractions, just in different get ups, and all, sort of speak, and vice versa, lol. So decimal powers are in essence and affect fractional powers. And, so, fractional or decimal powers are simply, what I call, don’t know if anyone else does, radical powers. “…what’s a radical power..!” This symbol, √, is called the radical. The radical is just another name for taking a specific root of a number, or for taking a specific root of a base, if you will (the two terms, number and base, are used interchangeably), such as, the square root, the cube root, the sixth root, etc., of a number. The base, or any expression beneath the radical symbol is called the “radicand.” The radical for the cube root of some number, or some base, if you will, will have a 3 written above and inside the little v shaped portion on the lower left side of the radical symbol. The 3, or root, is called the “index.” So, the radical symbol for the 6th root of some base (“radicand”) will have 6 written in along with the radical symbol. And so on and so forth for all the other roots, or “indexes,” taken of numbers, or bases, or “radicands,” if you will. But the square root radical symbol is the only exception to that rule. The square root radical symbol has an “index” 2, but the 2 is never written in along with the radical symbol but is always understood and accepted to be there, nonetheless. So only it alone, the square root radical symbol, looks like this, √, all the time.. never no 2! So… for instance, hopefully now that that’s settled, for the 5th root of a base the radical symbol will have its “index” 5 along with it, and likewise for all the other roots with respect to their “indexes,” again, except for the index 2 for square root of a number. For the 19th root of a number, for instance, 19 will be the “index” written in along with the radical symbol.

[@ndailorw5079]

@orangebetsy3404 So… now, to answer your question, when we take the square root of some number, we’re in effect raising that number to the power of 1/2! Which by another name and method we simply call raising some number to some power, or, doing exponents! So, in doing 2^1.6, or 2^(8/5), we’re still simply doing exponents, inherently! Just can’t do them the traditional way, as done when doing “normal” exponents, lol. For example, The √ 4 = 4^(1/2). Which simply means the square root of 4 raised to the power of 1, or, the square root of 4 raised to the 1st power! The missing “index” 2 being understood as being there without writing it. The missing 2 in the index notation is simply the accepted and agreed upon convention for expressing the square root of a number. Careful though! The “radicand”, or the base, which is the same thing, if you will, must always be raised to the specified power attached to it before taking the root of the result of that calculation! Next, for the √4, for example, √4 = 4^(1/2). The numerator 1 of the fractional exponent is the power to which the base is to be raised, and denominator 2, the “index,” is the root we’re taking of the base! It would look like this in long hand, √(4^1), with the missing “index” 2 being understood as being with the radical symbol, or, it’s implied, sort of speak. So, we have the square root of 4 raised to the power 1, or, 4^(1/2)! In other words, first raise the base to the power specified by the numerator of the fractional exponent, then take the root, as specified by the denominator, of that result ! Likewise for all other powers and roots.

[@ndailorw5079]

@orangebetst3404 So, for your 2^1.6 or 2^(8/5), we’d simply take the 5th root of 2 [(5), the denominator, is the root!] raised to the 8th power [(8), the numerator, is the power 2 is to be raised to!]. Or better and even safer yet, and more properly and correctly, we “must always” “first” raise 2, the base, or the “radicand,” pick your poison, to 8th power, then we take the 5th root of that result! And that’s it! After all, you must remember, decimals and percentages are simply fractions in another form, or outfit, and vice versa! But… try it on your calculator. Raise 2 to the 8th power, then take the 5th root of that result, and you should get 3.03143, rounded to 5 decimal places, for your answer. But that particular answer will just be the result and answer for your particular question of how to raise 2 to the 8/5 power, or how to raise a base to a fractional power. That result won’t be the answer to the problem the instructor presents here, of course. But you must make the inputs into your calculator properly or you’ll get a wrong answer, or an error message! But just know and remember that when doing bases raised to decimal powers the decimal must first be changed to a fraction to put the terms of the expression into their fractional form in order to “first” raise the base to the power specified by the numerator, then take the root of the base as specified by the denominator. Otherwise, if our fractional power remains in decimal form.. well… that’s then one reason why we have logarithms and exponentials and logarithm and exponential tables and slide rules, etc.… and calculators! Hope that helps more than it hurts…? …and hope you’re not winded after such a long wind….?

[@ndailorw5079]

@orangebetsy3404 …on second thought, and my apologies, guess I could’ve simply said, in a word or two, without all the long-winded foundational and fundamental preliminaries, simply change the base’s exponent, which is expressed in decimal notation, to a fraction, and from there the numerator is the power to which the base “must first” be raised before taking the root, and the denominator is the root to be taken of that result….? For another example, x^(0.27) ≈ X^(3/11) = the 11th root of x to the 3rd power, but more correctly, the 11th root of x after x has been raised to the 3rd power, and not before that step… and you’re done! Again, the √25 = 25^(1/2). 2 is the root to be taken of 25, and 1 is the power to which 25 must first be raised before taking the root!

[@benquinneyiii7941]

Everything in math is the opposite of something

[@The54Adrians]

I am 69 great to redo and learn arithmetic

[@abc17208]

I solved it fast without using a pen and paper. The answer is: x=2/5

[@abc17208]

8/5

[@1961ebutuoy]

x = 7/5

[@brawnerbandwidthmusic2529]

People say I’m smart, but untrained. It’s gotta be true. Self teaching yourself math is harrrrrddd or RDRR

[@user-vh3st3fw3c]

X=8/5

[@1961ebutuoy]

Ooops! I failed counting. x = 8/5

[@argonwheatbelly637]

8/5 in about eight seconds. Sorry, too long. 😔

[@patwilliams625]

And dont want to

[@billi8927]

Oral question😑

[@harrymatabal8448]

My advice to the aithor to change his negative attitude, it is very discouraging and mean. Ot is reflection of stupidity

[@b213videoz]

first thing to do is to express 16 as 2⁴ then it's doable: 2^(x-1) 2⁴ -------------- = ------; 2^(3-4x) 1 2^(3-4x) * 2⁴ = 2^(x-1); 3-4x+4 = x-1 8 = 5x x = 8/5 = 1⅗

[@k.t.ceducation2069]

X=8/5