Find number of solutions to the equation log

Question

Find number of solutions to the equation $\log_{x+1}{(x-\frac{1}{2})}=\log_{x-\frac{1}{2} }{(x+1)}$

The only solutions in the complexes are 1, and 3/2?

Let $(x+1)^a=(x-\frac{1}{2})$ and thus $(x-\frac{1}{2})^a=(x+1)$. So $(x+1)^{a^2}=x+1, (x-\frac{1}{2})^ {a^2}=x-\frac{1}{2}$. Now, if at least one of $x+1, x-\frac{1}{2}$ isn't a primitive root of $1$, then $a^2=1$. If $a^2 \neq 1$, it means $(x+1)^{a^2-1}=(x-\frac{1}{2})^{a^2-1}=1$ as obviously they are nonzero. Thus, if $p+iq$ for reals $p,q$ then $|x+1|=|x-\frac{1}{2}|=1$ and thus $(p+1)^2+q^2=1, (p-\frac{1}{2})^2+q^2=1$ so $p=-\frac{1}{4}, q=\pm \frac{\sqrt{7}}{4}$, one can verify that none of the two are primitive roots of $1$. Thus we come to the other case $a^2=1$. Obviously $a$ is not $1$ as it would mean $x+1=x-\frac{1}{2}$. So it means that $a=-1$ and so $(x+1)(x-\frac{1}{2})=1$ or $x^2+\frac{x}{2}-\frac{3}{2}=0$ or $2x^2+x-3=0$. The roots are $x=1, x=\frac{3}{2}$ and they are the only solutions over complex numbers.

Answer 1

We have that

$$\log_{x+1}{(x-\frac{1}{2})}=\log_{x-\frac{1}{2} }{(x+1)} \iff \frac{\log{(x-\frac{1}{2})}}{\log {(x+1)}}=\frac{\log {(x+1)}}{\log{(x-\frac{1}{2})}} $$

$$\iff \frac{\log{(x-\frac{1}{2})}}{\log {(x+1)}}=\pm1 \iff \log{(x-\frac{1}{2})}=\pm (\log {(x+1)})$$

that is

$$x-\frac{1}{2}=x+1\qquad \lor \qquad x-\frac{1}{2}=\frac1{x+1}$$

with

• $x+1>0 $ and $x+1\neq 1 $
• $x-\frac{1}{2}>0$ and $x-\frac{1}{2}\neq 1$