Exponential equations involving natural numbers at power "x" Find x : $$4^x+15^x=9^x+10^x(2^x-3^x)(2^x-3^x-5^x)$$

Oops - I made a mistake:
$x=0$ is NOT a root
since
$3^0-2^0 = 0$, not $1$.
Please retract my points.
I'll write it as
$$4^x+15^x=9^x+10^x(3^x-2^x)(5^x+3^x-2^x)$$
We see that $x=0$ is a root,
since
both sides are $2$
(because all the $n^x$ are 1).
(No it's not, but I talked fast, didn't I?)
I'll submit this and then look for more.
If $x=1$,
the two sides are
19 and 69, so the right side is larger.
If $x \ge 1$,
the right side is at least
$9^x+10^x 5^x
= 9^x+50^x$
and this is larger than the left side.
For the fun of it,
I'll try $x=-1$.
The left side is
$\frac1{4}+\frac1{15}
= \frac{19}{60}
$
and the right side is
$\frac1{9}+\frac1{10}(\frac1{3}-\frac1{2})(\frac1{5}+\frac1{3}-\frac1{2})
= \frac1{9}+\frac1{10}(-\frac1{6})(\frac{6+10-15}{30})
=\frac1{9}-\frac1{10}\frac1{6}\frac{1}{30}
=\frac1{9}-\frac1{1800}
$.
These are not equal, so -1 is not a root.
I'll stop here.
