Exponential equations involving natural numbers at power "x"

Question

Find x : $$4^x+15^x=9^x+10^x(2^x-3^x)(2^x-3^x-5^x)$$

Answer 1

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.