# Why do logarithms produce such difficult problems

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.

• 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. Jan 16, 2014 at 18:59
• Are logs really special in producing difficult problems? Even a fifth degree polynomial can be pretty tough. Jan 16, 2014 at 22:03
• 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 $\log_{a}\left(a\right)=1-\log_{b}\left(1\right)$, $1=1-0$, $1=1$.
– JAB
Jan 17, 2014 at 13:12

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.

• 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. Jan 16, 2014 at 19:25

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 $$\log_5(x+3)=1-c\log_5(x-1)\\ \Updownarrow\\ \log_5(x+3)+c\log_5(x-1)=1\\ \Updownarrow\\ (x+3)(x-1)^c=5$$

Here is another $\log$-equation, you could find entertainment in solving: $$\log_5(x)+\log_{25}(x)=\log_{5}(\sqrt{3}x)$$
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.
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 $x=2$.