Exponential equation on the set of real numbers

Question

Solve the following equation on the set of real numbers: $8^x+27^x+2·30^x+54^x+60^x=12^x+18^x+20^x+24^x+45^x+90^x$

$x=1; x=0; x=-1$ are trivial solutions, but I'm stuck with proving that there are no others...

Answer 1

Assuming we are looking for real valued solutions for $x$:

Start by writing all the terms on the left hand side, with zero on the other side, and write $a=2^x$, $b=3^x$ and $c=5^x$

The equation then becomes $$a^3+b^3+2abc+ab^3+a^2bc-a^2b-ab^2-a^2c-a^3b-b^2c-ab^2c=0$$

When $b$ is replaced by $a$ the polynomial is identically zero, so $(b-a)$ is a factor.

So we have one solution already, given by $$b-a=0\Rightarrow 2^x=3^x\Rightarrow x=0$$

After some long division, the remaining factor is $$a^2b+ab^2-abc-a^2+ac+b^2-bc$$

This factorises as $$(ab-a+b)(a+b-c)$$

So there are two more possible solutions to consider:

firstly,$$2^x3^x-2^x+3^x=0\Rightarrow 6^x+3^x=2^x,$$ which has only one solution, namely $x=-1$

And secondly, $$2^x+3^x-5^x=0$$ which again has only one solution, namely $x=1$

So the three solutions you identified in the first place are indeed the only solutions.