Solve the equation $ \log_2(9x+2)=\log_3(16x+3). $ 
Solve the equation
  $$
\log_2(9x+2)=\log_3(16x+3).
$$
  It is easy to see that $x=0$ is a solution but how to prove that there are  no more solutions?

My idea was prove that the function $f(x)=\log_2(9x+2)-\log_3(16x+3)$ is monotone for $x \geq 0$. We have 
$$
f'(x)=\frac9{ \left( 9\,x+2 \right) \ln  \left( 2 \right) }-
\frac{16}{ \left( 16\,x+3 \right) \ln  \left( 3 \right) },
$$
and the inequality $f'(x)>0$  has the solution 
$$
 \left( -2/9,  -3/16 \right)  \cup 
  \left( a 
 ,\infty  \right), 
$$
where $$a=-{\frac {1}{144}}\,{\frac {32\,\ln  \left( 2 \right) -27\,\ln 
 \left( 3 \right) }{\ln  \left( 2 \right) -\ln  \left( 3 \right) }}$$
so $f'(x)>0$ for $x>0$. Thus $f(x)$ is there monotone and $f(x)=0$ has only one solution for $x \geq 0$.
My question  Is there a solution without using the derivative?

YES, we can solve this equation without using derivative.
 here I solved this equation by a method without using derivative.
$\log_2(9x+2)=\log_3(16x+3)=y $(say)
$\log_2(9x+2)=y \implies2^y=9x+2$
multiply this equation by 16$\implies2^{y+4}=144x+32$.........(1)
second part of given equation is $\log_3(16x+3)=y\implies 3^y=16x+3$
multiply this equation by 9 $\implies3^{y+2}=144x+27$.........(2)
$(1)-(2)\implies 2^{y+4}-3^{y+2}=5$.......(3)
obviously $y=1,-1$ is a solution
if $y=1\implies\log_2(9x+2)=\log_3(16x+3)=1$.By using the formula: if,$log_a(b)=1\implies b=a.$
similarly $y=-1$ we find $x=-1/6$
$9x+2=2\implies x=0$. 
similarly, $16x+3=3\implies x=0$
$x=0,-1/6$ is a solution . we want to prove $x=0,-1/6$ is only two solution
from (3),  $2^{y+4}-3^{y+2}=5$,
if $y$ is greater than $1$, $2^{y+4}-3^{y+2}$,is negative.i.e,$2^{y+4}\lt3^{y+2}$.we prove it by induction
it is true for $y=1$,let it is true for $y=k$.
i.e,$2^{k+4}\lt3^{k+2}$ we know that $2\lt3$
multiply these two inequalities,we get $2^{(k+1)+4}\lt3^{(k+1)+2}$, by $PMI$ it is true for all natural numbers greater than $1$.
for $y\gt1$,there is no solution for$2^{y+4}-3^{y+2}=5$,since , $2^{y+4}-3^{y+2}$ is negative for $y\gt1$.
 therefore, there is no corresponding value for $x$.
Therefore $\log_2(9x+2)-\log_3(16x+3)$ has no solution other than $x=0,-1/6$
