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This was inspired by Fun logarithm question, because it made me remember a question I accidentally asked on a quiz some time ago. It was suppose to have both log bases the same, 3 or 5. $$\log_{5}\left(x+3\right)=1-\log_{3}\left(x-1\right)$$ After apologizing to my students, I talked to some people about it and we could not find an analytical solution... other than to realize that $x=2$ is a solution by just trying it. So, my questions are: Is there an analytical solution to this specific problem? And, more importantly, why do variables in exponents/logarithms that are seemingly easy to state produce such difficult problems? I would like some insight into the second question more than the first, as answering the second will also answer the first, I think.

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  • $\begingroup$ Mathematica can solve it to give $x=2$ as the exact solution, but it looks like it can't do more general examples without resorting to Root[] objects. $\endgroup$ Jan 16, 2014 at 18:59
  • $\begingroup$ Are logs really special in producing difficult problems? Even a fifth degree polynomial can be pretty tough. $\endgroup$
    – Jack M
    Jan 16, 2014 at 22:03
  • $\begingroup$ As an extension, the reason the integer 2 (as opposed to a non-integer real value) is an exact solution is that this is a case of the more general equation $\log_{a}\left(x+u\right)=1-\log_{b}\left(x-v\right)$ where $x-v=1$ and $x+u=a$, which reduces the problem to $\log_{a}\left(a\right)=1-\log_{b}\left(1\right)$, $1=1-0$, $1=1$. $\endgroup$
    – JAB
    Jan 17, 2014 at 13:12

4 Answers 4

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If you put everything in a common base, let's say 5, this equation is equivalent to $$(x+3)(x-1)^c=5 $$, with c = $\log_53 \approx 0.68261$. Solving expressions with polynomials is usually easy. I would guess that the problem here is that the exponents are not integer, but real, which makes everything harder.

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  • $\begingroup$ I guess logs only give the illusion that something really complicated is in fact simple. For example, at first sight I have no idea how much $5^{0.5693}$ is, or even if it's closer to 0 or 5. But if I was told that it is almost the $\log_5 2.5$, things would seem easier. $\endgroup$ Jan 16, 2014 at 19:25
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Ad question 1

Like user120820 already indicated, you can solve it analytically by converting into a common base via the change of base formula $$ \log_b(x)=\frac{\log_k(x)}{\log_k(b)} $$ With this we have $\log_3(x-1)=\frac{\log_5(x-1)}{\log_5(3)}$. To simplify things we define a name for the constant factor $c=1/\log_5(3)\approx 1.46497352072$ that it costs converting a $\log_3$ to a $\log_5$. Then the equation becomes $$ \log_5(x+3)=1-c\log_5(x-1)\\ \Updownarrow\\ \log_5(x+3)+c\log_5(x-1)=1\\ \Updownarrow\\ (x+3)(x-1)^c=5 $$

Ad question 2

The problem you gave involved only addition and multiplication along with different bases of the logarithms involved. The change of base part produced an extra step that was difficult to identify and deal with.

Because the rules for logarithms are so suprising in a way, it is a bit like solving a puzzle where the steps in the puzzle are so entangled with themselves, that it becomes complicated to keep track of your moves...

Here is another $\log$-equation, you could find entertainment in solving: $$ \log_5(x)+\log_{25}(x)=\log_{5}(\sqrt{3}x) $$

Regards, String

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The whole idea is, after noticing that $x=2$ is a solution, to take the derivative, and observe that it is strictly positive for $x\ge1$, meaning that the function itself is monotonously increasing on $[1,\infty)$, hence the solution is unique.

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Also not really an analytical solution, but another way to "visualize" the solution is to make a substitution $$3^t=x-1$$ then the equation transforms to $3^t+4=5.5^{-t}$ or $$3^t5^t+4.5^t=5$$ which is the same as $$(4-1)^t(4+1)^t+4(4+1)^t=1+4$$ Then it is clear that if $(4-1)^t(4+1)^t=1$ and $(4+1)^t=1$ then we have solution. Those two equations hold true for $t=0$ and plugging it back to the original substitution, we obtain $x=2$.

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