how to solve logarithmic equations with different bases

Solving logarithmic equations with different bases However, I do not know the approach to getting the solution to one of them.
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#math #algebra #mathbasics #logarithm

[@stevemonkey6666]

It says something good about you that you posted a problem that you cannot solve. But I cannot solve it either. 😔

[@namelessnormie]

so here is a simple mathematical approach to solve: log_(2)x=log_(3)(x+1) 1) use change of base rule(as mr BPRP) (lnx )/(ln2) = (ln(x+1))/(ln3) then rewrite the equation:- (lnx )/(ln(x+1))=(ln2)/(ln3) now we know two things 1) lnx=ln2 and 2) ln(x+1)=ln(3) we know log_(a)b = log_(a)c if and only b=c so based on this x=2 and x+1=3 ........x=2 but now the problem is this approach will not work if the question was log_(2)x= log_(5)(x+1)

[@michaelempeigne3519]

log_2 ( x ) = log_3 ( x + 1 ) = y so 2^y = x and 3^y = x + 1 3^y = (2^y ) + 1 3^y - 2^y = 1 this is easy to tell that y = 1 that means log_2 ( x ) = 1 so x = 2

[@tianqilong8366]

sheeesh, brilliant!

[@udic01]

How is it easy to tell? I mean i know that it is 1, but how to prove it?

[@michaelempeigne3519]

@@udic01 if you really want to prove it, you can do it with some number theory

[@tianqilong8366]

@@udic01 you can try this i think: Claim that y = 1 is the only solution. When y > 1, the derivative of 3^y is bigger, so 3^y and 2^y deviates further and further, which means 3^y - 2^y > y(1) = 1, which shows that the claim is valid when y > 1. Similarly, when y < 1, since 3^y has a steeper slope, the gap between the two curves 3^y and 2^y closes out, 3^y - 2^y < y(1) = 1, so the claim is aldo valid when y

[@andrewelmquist6829]

@@udic01 Okay this is following someone else's proof. log_2(x)=log_3(x+1) 2^y=x 3^y=1+x 3^y=1+2^y add one so we dont divide by zero later down the road. 3^y+1=2^y+2 okay so. This is where it gets. Weird. (3^y)^1/y=3 This next step is important. Taking into account that 2^y=x, if y were to equal 0, then we would get 1=x. This would then mean that 1+1=3^0, which also means that 1+1=1. This is false, and means that y CANNOT equal zero. (1)^(1/y)+(3y)^(1/y)=(2^y)^(1/y)+2^(1/y) since y does not equal 0, because of the fact that 1^n (any real number, decimal, negative, positive, ANYTHING)=1. This is only not true is n=0, but since y does not equal zero, we do not need to worry about that, due to us already ruling out that it is false. That being said, that means 1^(1/y)=1. Rewrite function with this in mind. Also then canceling out all the power of y raised to 1/y, because that means 1. (rooting any number with a power equal to the index just equals said number). 1+3=2+2^(1/y) 4=2+2^(1/y) 2=2^1/y Take the ln (or any log lol) on both sides. It doesnt matter because it will cancel out. ln(2)=(1/y)*ln(2) DIVIDE ON BOTH SIDE COME ON LAST STEP ln(2)/ln(2)=1/y 1=1/y multiply on both sides. Finally. 1y=1. y=1 plug this back in. 2^y=x 2^1=x x=2 QED. Dont ever ask me to do this again.

[@EE-ho1iz]

My solution is as follow: log_2(x) = log_3(x+1) = ln(x)/ln(2) = ln(x+1)/ln(3) (as shown in the video) Here's where I deviate from Steve's path: I multiply both side by ln(2)/ln(x+1), which yields: ln(x)/ln(x+1) = ln(2)/ln(3) By comparison, x=2 and x+1=3, which of course still yields x=2 Therefore x=2. Voila!!! Edit: An honorable mention to Chirag Ahuja in the comment section, who came up with e^ln(x) = e^ln(2)log_3(x + 1) through some manipulation, and had the idea of "compare ln(x) with ln(2) and log_3(x + 1) with 1". It also used the method of solving by comparison. It's so interesting that I can't help but mentioning it!

[@ieatgarbage8771]

Huh, so our question becomes “for what value k will 2^k+1=3^k,” and x=3^k. That seems like a much more familiar. When I plug similar problems into wolframAlpha (such as 3^x+1=5^x), they don’t even have an exact answer, only approximations. This tells me that no one’s figured out how to solve these but also no one cared enough about these to give them a function. Maybe there is a general solution out there somewhere. Here’s an important question: how do we know lambert W is non-elementary?

[@chiragahuja2309]

I appreciate the mention, thanks a lot.

[@chiragahuja2309]

So these can be the possible explanations: 1. So I ended up with a equation after some manipulations e^ln(x) = e^ln(2)log_3(x + 1) my thought is how about after equating the exponents since bases are same we compare ln(x) with ln(2) and log_3(x + 1) with 1 this works out both yields the ans as 2 and makes sense as well. It might be wrong but makes sense for me. 2. another expression is ln(3)^ln(x) = ln(x+1)^ln(2) now we can just compare ( ln(3) = ln(x+1) ^ ln(x) = ln(2) ) and we end up with 2 3. One more I came up with is: Log_(x+1)(x) = log_3(2) Am thinking of comparing bases and the number which we are taking log of but this just seems wrong but on a second thought I thought I should still put

[@WolfgangKais2]

Transforming the second equation to ln(x+1)/ln(x) = ln(3)/ln(2) helps to „see“ the solution, and showing that the LHS is decreasing (for x>1) ensures that there is only one solution. I was able the solve the related equathions when replacing the log base 3 by base 4 or base 8, and I am pretty sure that there is no solution when replacing base 2 by base 5.

[@moskthinks9801]

I don't know how to solve the case for different bases, but I can show this equation only has x=2 as its answer. Note that it satisfies the equation. Then define f(x)=log2(x)-log3(x+1) We have f'(x) = 1/(xln2)-1/((x+1)ln3) = (xln(1.5)+ln(3))/((x^2+x)ln(2)ln(3)) So for the domain of f(x) [x>0], we have f'(x)>0, which means f is strictly increasing. Because of that, f(x) must have only 1 root, which happens to be x=2. QED. Although now what😅

[@tmsniper9229]

here's what i thought: ln(x)/ln(2)=ln(x+1)/ln(3) which gives ln(x)/ln(x+1)=ln(2)/ln(3) if we call the function on the left side f(x)=ln(x)/ln(x+1) then we are looking for solutions of f(x)=f(2), it's sufficient to show that f is one to one in order to prove that x=2 is the unique solution, f'(x)=((ln(1+x)/x)-(ln(x)/(x+1)))/(ln(x+1))² =((1+x)*ln(1+x)-xln(x))/ln(x+1))²>0 for all x>0 so we can safely say that f is increasing, therefore it is one to one so x=2 is the unique solution.

[@revrift6447]

What a chad

[@nicolascalandruccio]

In fact, most cases are not explicitely solvable. The general case is: log_a(x)= log_b(x+1) where, x>0, a>1, b>1 and b>a Let t=ln(b)/ln(a) Thus, t>1 ln(x)/ln(2)=ln(x+1)/ln(3) (ln(3)/ln(2)).ln(x)=ln(x+1) t.ln(x)=ln(x+1) ln(x^t)=ln(x+1) x^t=x+1 x^t-x-1=0 Therefore, there’s no explicit solution even using Lambert W function except in some rare cases where a is a natural number and b=a+1 solvable by comparison and the solution is x=a as @Chirag Ahujalike said for the case where a=2 and b=3 or some polynomial cases where t=2, t=3 or t=4. For instance, the solution for t=2 is the golden ratio and for t=3, x=cbrt((9+sqrt(69))/18)+cbrt((9-sqrt(69))/18)~1.32472

[@nicolascalandruccio]

Oh! I've just realized it means: for all natural numbers n^(ln(n+1)/ln(n))=n+1

[@padraiggluck2980]

For the second problem, multiply the first line in blue on both sides by ln(3)/ln(x) to get ln(3)/ln(2) = ln(x + 1)/ln(x) and the answer x = 2 is obvious.

[@padraiggluck2980]

@@ensiehsafary7633 Of course it was obvious. The point of the demonstration was to prove it. bprp’s proof can be reduced to a line or two

[@ZipplyZane]

I like this one the best. Seeing you have the same functions on both sides tells you to examine if the inputs are also the same. So you set 3 = x+1 and 2 = x, and see that those both produce the same answer. Hence you know that x=2 is a solution. From there, it just remains to show that there is only one solution.

[@好吧-h6k]

I’m still trying to solve the problem algebraically, I’ve literally tried every substitution

[@jabarzua]

But you're still solving by inspection. The problem is to algebraicly get to x=2

[@padraiggluck2980]

@@jabarzua First of all, who said the problem must be solved algebraically? The problem involves logarithms, it is nonlinear. Second, you were not paying attention to the video: the log is bijective which means that when you arrive at log(x) = log(2), regardless of the base, you can conclude that x = 2.

[@kitsunetheedgeofnightfall]

if the bases are equal, can't he just equate the results? it would then become x/2=x+1/3... 3x=2x+2... 3x-2x=2... x=2?

[@tbg-brawlstars]

Let base be a after we use base change property Then 1st answer is x = a^[(log3log2)/(log3log2)] all logs to base a 2nd one was easier than one Here's how to do it Logx/log2 = log(x+1)/log3 Using base change property reverse.... Log3/log2 = log 3 base 2 and log (x+1) /logx = log(x+1) base x Log3 base 2 = Log (x+1) base x X+1 = 3 X= 2 By comparing eqn. x=2

[@Mediterranean81]

Lambert W

[@olzenkhaw]

Just solve it by graph.

[@BoagoGopadile]

Solve log2(27)=x² - log81(8)

[@AaronWGaming]

Umm going back a few steps what if you Divide both sides by the Fractions to FLIP the Fractions

[@AaronWGaming]

If both sides are equal Ln2=Ln3 OR Ln(X) since it can't be Ln3 you get Ln(X)=Ln(2) therefore X=2

[@ayxd1386]

Hi, can anyone please explain why was there a 1 on the other side rather than a 0, @ 4:28

[@andrewelmquist6829]

Okay this is following someone else's proof. log_2(x)=log_3(x+1) 2^y=x 3^y=1+x 3^y=1+2^y add one so we dont divide by zero later down the road. 3^y+1=2^y+2 okay so. This is where it gets. Weird. (3^y)^1/y=3 This next step is important. Taking into account that 2^y=x, if y were to equal 0, then we would get 1=x. This would then mean that 1+1=3^0, which also means that 1+1=1. This is false, and means that y CANNOT equal zero. (1)^(1/y)+(3y)^(1/y)=(2^y)^(1/y)+2^(1/y) y does not equal 0, because of the fact that 1^n (any real number, decimal, negative, positive, ANYTHING)=1. This is only not true is n=0, but since y does not equal zero, we do not need to worry about that, due to us already ruling out that it is false. That being said, that means 1^(1/y)=1. Rewrite function with this in mind. Also then canceling out all the power of y raised to 1/y, because that means 1. (rooting any number with a power equal to the index just equals said number). 1+3=2+2^(1/y) 4=2+2^(1/y) 2=2^1/y Take the ln (or any log lol) on both sides. It doesnt matter because it will cancel out. ln(2)=(1/y)*ln(2) DIVIDE ON BOTH SIDE COME ON LAST STEP ln(2)/ln(2)=1/y 1=1/y multiply on both sides. Finally. 1y=1. y=1 plug this back in. 2^y=x 2^1=x x=2 QED. Dont ever ask me to do this again. (THIS PROOF IS WRONG, EXPONETIAL STEP IS INCORRECT, BUT NEW PROOF WAS MADE IN COMMENTS)

[@andrewelmquist6829]

Okay so, that proof was wrong. Lol. Made an easy mistake and only got the right answer because y=1. But i have the right stuff. Proof: 3^y=2^y+1 Move 2^y over with 3^y 3^y-2^y=1 We then root it to t he index of y (the step i messed up) yrt(3^y-2^y)=yrt(1) The logic used t prove y is not 0 and that 1 to any real power is 1 is true. Same logic there. yrt(3^y-2^y)=1 Now. Rewrite it for the ln function (3^y-2^y)^1/y=1 Ln both sides. (1/y)(ln(3^y-2^y)=ln(1) okay, so before we go any further add 1 to both sides to we dont divide by zero. At the same time, im taking taking the ^y and bringing to the front, a simple rule of logs y*y*1/y(ln(3-2)+1=ln(1)+1 I am going to sellectively calculate the lns, and this is so i dont have to divide by zero or so some terms dont just cancel out. Im also getting rid of the mess of ys at the beginning, due to the fact 2 cancel out and we are just left with a y y*(ln1)+1=1 divide by y ln(1)+1=1/y ln(1)=0 0+1=1/y Final. Its right. Multiply on both sides. y=1 QED

[@tsdtech1065]

Answer for the 1st one is nearly 6.542

[@seblara839]

I just love math, I don't even study it but I think I got something, not it but something from lnx/ln2=ln(x+1)/ln3 multiply both sides by ln3/lnx and get ln3/ln2=ln(x+1)/lnx and by laws of logarithms log_2(3)=ln_x(x+1) I did not get that X=2 BUT I did get that. please feel free to comment your thoughts Hi from Mexico;)

[@abubakrsayfullayev9003]

Man I knew the asnwer is two before you tell the answer

[@juliang8676]

Ln(x)/ln2=ln(x+1)/ln3 Ln(x)*ln(3)/ln(2) =ln(x+1) Ln3/ln2=a. Note a>1 x^a=x+1 F(x)=x^a-x-1 F'=x^(a-1)-1=0 at x=1 F''=x^(a-2)>0 at x=1 therefore x=1 is the global maximum of F(x) (F'=0 has only one solution). To find root analytically i have no idea, thought there might be smth here.

[@hrcsk2479]

Then with newton method we can say x is approaching to 2 so x=2

[@tsa_gamer007]

Log(base2)X=Log(base(2+1))x+1 By comparison x=2

[@HarronsLife]

I think we can solve it this way Let log2(x)=log3(x+1)=a. then log2(x)=a, log3(x+1)=a. then x=2^a and x=3^a-1 then 2^a =3^a-1 then 3^a-2^a=1 then 3^a-2^a= 3-2 then 3^a-2^a=3^1-2^1 then a=1 thus sub back: log2(x)=a=1 or x=2^a=2^1 or others. then x=2. BINGO HAHAHAH CAN WE SOLVE THIS QUESTION USING THIS WAY?

[@serulu3490]

Thankfully i solved this despite not learning log in school. Log really seems easy, it's rules make a lot sense

[@purple2098]

How tf

[@OptimusPhillip]

First equation: x = e^(1/ln(3/2)) [Well, I got my properties of logs mixed up. Somehow, I got the idea that 1/ln(a) = -ln(a)] Second equation: I would know how to solve it if there was an identity for products/quotients of natural logs. As it sits, I'm stuck at ln(x)/ln(x+1) = 1/(ln(2)*ln(3)) [How the hell did I forget that ln(2) was in the denominator of the initial expression? I was on the verge of greatness, I was this close!]

[@jessetrevena4338]

I love you videos!! I watch every time you upload :)

[@davidseed2939]

I think we are being asked for s method, that does not rely on an essy answer. eg log_2(x)=log_5(x+1) log_2(x)=log_2(x+1)/log_2(5) log_2(x)=a.log_2(x+1) x=(x+1)^a x^(1/a)-x=1 exp(x^b)/exp(x)=exp(1)=0 oops

[@julienmcat]

Here’s the solution to the second question: log2(x) = log3(x+1) log(x) / log(2) = log(x+1) / log(3) Use cross substitution: log(x) / log(x+1) = log(2) / log(3) According to the equation, the numerators and denominators of both sides equal each other. Use the numerators to solve for x: log(x) = log(2) Therefore x = 2 You can also use the denominators to solve for x: log(x+1) = log(3) Therefore x = 2

[@jabarzua]

Just make x->0

[@williamm2127]

using ln(x+1)/ln(x) = ln(3)/ln(2) you can state x+1:x=3:2 then (x+1)/3=x/2 and you can solve from there

[@twocsies]

x^-ln2 = (x+1)^-ln(2+1) && f(x)=x && g(x)=x+1 => f(x)^-ln(f(2)) = g(x)^-ln(g(2)) so by the power invested in transcendental numbers, if we set both sides equal to e, then x=2.

[@yakupbuyankara5903]

X=2

[@betibeli7319]

log6^x=log3^4x

[@lexiette8343]

I think I know how to prove the latter equation, rewriting the original to: lg(x) / lg(x+1) = lg2 / lg3, you can then let 2 be y instead and now you have the same on both sides except for that x is y, and since its all in logs that means x=y!

[@joeanderson2024]

I somehow got the second equation to equal ln(3)/ln(2)=ln(x+1)/ln(x) and figured it would make sense to set the inside of the numerators and inside of the denominators equal getting x+1=3, or x=2, and x=2

[@TheRodrigo351]

by comparision was the way everyone resolves the equation in the head: log2(x) = log2+1(x+1) -> x=2 done.

[@neomooooo]

So we’re looking at log_a(x)=log_b(x+c), as a general rule. Or, if we change notation, m*ln(x)=ln(x+c), with m=ln(b)/ln(a). This leads to the nonlinear equation x^m=x+c, which will only have a positive real solution if either m>1 and c>0, (a line with a positive intercept intersecting a concave polynomial through the origin), or when m

[@saharhaimyaccov4977]

We need to find some power that 3^x=2^x ... Maybe in imaginary num

[@michaeledwardharris]

Well that was a first haha. Super interesting problems though! Nicely done, as always.

[@MASDOOD_]

الجبر= Algebra

[@ruilongsheng2845]

log x (1/log2 -1/log 3)=1, log x= log 2*log3 /(log 3-log 2)

[@abrahammekonnen]

Yeah that 2nd equation is a monster.

[@abdoshaat3304]

Why switch from log to ln

[@iyousef46]

Is there any benefit to using (ln) instead of (log base 10)? We didn't learn about (e) or (ln) yet in school but solved similar questions to your first example using (log10).

[@leochinchillaa]

for this you could use any log base you want, even log base 57. you will still arrive at the same answer. i guess most use ln(b) because it s a lot easier than writing out log base (a) of b (with proper notation of course)

[@persivald2148]

x = exp ( ln 3 / log2 (1.5))

[@thrvfty]

I had a similar problem and I haven't been able to solve it. The problem is a system of equations as follows: x = y^2 and y = 3^x - 7. For the life of me, I cannot figure out the answer and looking online doesn't seem to help. If anyone has an idea on how to solve it, please tell me.

[@frankinfa9212]

I believe you should use a numerical method such as Newton or bisection method. You can solve y=3^(y^2)-7 respect to y, and then evaluate x. Just imagining the graph of the equations, there should be 2 points of interesection, and so, the system has 2 solutions. Hope this is usefull.

[@thrvfty]

@@frankinfa9212 Thanks for your suggestion. I solved the equation through graphing already, but I was wondering if there was some algebraic solution, not just an approximation. I'm going for an elegant solution rather than a practical one.

[@Earlforsure]

Ermmm can someone help me with my school circle question....my teacher gave me the question but she also didn't manage to get the answer....i bet this is not a simple or basic question

[@Earlforsure]

The question is: There is one straight line and have three circles on it(the line is the tangent for the three circles) The 1st circle>radius=4 cm The 2nd circle>radius=2 cm And in the middle of the two circles is the smallest circle that the question ask for the radius Lastly,all of the three circles connected to each other

[@MrGnome-ng6jv]

How long is the line?

[@MUJAHID96414]

I can solve

[@Armytechrex]

How?

[@MUJAHID96414]

@@Armytechrex if I try

[@Armytechrex]

@@MUJAHID96414 try and tell me

[@MUJAHID96414]

@@Armytechrex ok

[@MUJAHID96414]

@@Armytechrex I have solved 🔥🔥

[@GodbornNoven]

ln(x) / ln(x+1) = ln(2)/ln(3) now we can just get rid of the ln's x/(x+1)=2/3 x=(2x+2)/3 3x=2x+2 x=2

[@oenrn]

That's not how ln works, you can't "just get rid" of it.

[@andrewelmquist6829]

@@oenrn you can due to having common bases and the natural log function being one to one means that if the bases are the same, and the function outputs are equal, we can get rid of the lns. For instance, lets take ln(x-2)=ln(3). The simplest and easiest PROVABLE (not just looking at it and seeing the solution) is to cancel out the lns. Since both sides of the equation have them, and they are equal, the output for x will be the same as x-2=3. Proving that this is valid. ln(x-2)=ln(3) raise both sides to e e^(ln(x-2))=e^ln(3) cancel out the lns and es, saying that they are inverse function of each other x-2=3. QED We have proved that we can just "cancel" out the lns, due to the fact that the bases are the same, meaning the inside values have to be the same as well.

[@oenrn]

@@andrewelmquist6829 That has 0 revelance to what OP wrote. e^(ln x / ln (x+1)) is NOT the same as x / (x+1). You can't simplify logs like that. Which was my point.