How to solve the logarithmic equation $\log_2 (1 + \sqrt{x}) = \log_3 x$ analytically $$\log_{2}{(1+\sqrt{x})}=\log_3x$$ I find $x=9$ solution of this equation (by graphing).  Is there a way to solve it analytically ? 
$$x=9 \to \log_{2}{(1+\sqrt{9})}=\log_39  \space \checkmark$$
Remark:this question belong to calculus 1 exam of polytechnic university of Tehran.
 A: Substitute $1 + \sqrt x=z$  so we have $x=(z-1)^2$
The equation becomes
$\log_{2}{z}=\log_3(z-1)^2$
$\log_{2}{z}=2\log_3(z-1)$
Convert in natural log
$\dfrac{\log 3 \log z}{2 \log 2}=\log (z-1)$
$\dfrac{\log 3}{\log 4}=\dfrac{\log (z-1)}{\log z}$
$z=4$ is the unique solution of this equation since $f(z)=\dfrac{\log (z-1)}{\log z}$ is  increasing for any $z>1$.
Indeed $f'(z)=\dfrac{-z \log (z-1)+\log (z-1)+z \log z}{(z-1) z \log ^2 }$ is positive when numerator $-z \log (z-1)+\log (z-1)+z \log z>0$ that means
$z> \dfrac{\log (z-1)}{\log (z-1)-\log (z)}$ which is verified for any $z>1$ as the RHS is less than $1$ 
Therefore 
$z=4\to x=(z-1)^2=(4-1)^2\to x=9$
which is the only solution of the equation 
Hope this helps
A: Let $u=1+\sqrt{x}$ so that $x=(u-1)^2.$  Then the equation is
$$\log_2 u = \log_3(u-1)^2 = 2\log_3(u-1) = y,$$
where $y$ is some number.  Then $2^y=u$ and $3^{y/2} = u-1.$  Then
$$3^{y/2} = 2^y -1 = 4^{y/2} - 1$$
or
$$4^{y/2} - 3^{y/2} = 1$$
so we conclude that $y/2 = 1$, so $y=2$, and $u=4$ and $x=9.$
Fun problem.
A: I don't get it. There is rather standard way to do it. Calculate the derivate of 
$$ f(x) = \log _2 (1+\sqrt{x}) - \log _3 x$$ and show that it is monotonic. Thus result (that is 9 is the only solution, which you ''guess''). 
A: $$1 = 1 $$ 
$$\log_a a = \log_b b $$ 
$$n\log_a a = n\log_b b $$ 
$$\log_a a^n = \log_b b^n $$ 
then $ \sqrt x + 1 =2^n$ ....
$ x =(2^n -1)^2$
$ 3^n =(2^n -1)^2$
So for $n=2m $ even it gives
$ 3^m =(4^m -1)$
$m = 1$ is a solution..
A: $\log_{2}{(1+\sqrt{x})}=\log_3x
$
is the same as
$\dfrac{\log(1+\sqrt{x})}{\log 2}=\dfrac{\log x}{\log 3}
$.
The logs can be to any base. It doesn't matter,
as long as they are all the same.
Cross-multiplying,
$\log 3 \log(1+\sqrt{x})
=\log 2 \log x
$
or
$\log((1+\sqrt{x}))^{\log 3 }
= \log (x^{\log 2})
$
or
$(1+\sqrt{x}))^{\log 3 }
=x^{\log 2}
$
or
$1+\sqrt{x}
=x^{\log 2/\log 3}
$.
If
$f(x)
=x^{\log 2/\log 3}-(1+\sqrt{x})
$,
then
$f(1) = -1$
and
$f'(x)
=c x^{c-1}-\frac12 x^{-1/2}
$
where
$c = \frac{\log 2}{\log 3}
\approx .6309
$.
Then
$\begin{array}\\
f'(x)
&=c x^{c-1}-\frac12 x^{-1/2}\\
&=\frac1{2\sqrt{x}}(2c x^{c-1/2}-1)\\
&\gt 0
\qquad\text{for } x > 1\\
\end{array}
$
Therefore
$f(x)$
can have at most one root
for $x > 1$.
Since $x=9$
is a root,
it is the only root.
