Exponential equation on the set of real numbers 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... 

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.
