Find $x$ in the exponential equation $3^x+4^x+5^x=6^x$

Question

$3^x+4^x+5^x=6^x(R:x=3)$

I try but I can't finish

$3^x+2^x~2^x+5^x=3^x.2^x$

$3^x\cdot2^x-3^x-2^x\cdot2^x=5^x$

Answer 1

Comment:

In addition to Dietrich's algebraic solution, you may also apply this elementary reasoning:

$4^x+5^x=6^x-3^x=3^x(2^x-1)=3 y$

this relation deduces that left hand side must also be a multiple of 3. it can be seen that only with odd x this is possible:

$x=1\Rightarrow 4^1+5^1=9= 3\times 3$

$x=3\Rightarrow 4^3+5^3=64+125=189=3\times 63$

etc. You can check theses powers and find that :

$4^3+5^3=189=27\times 7=3^3(3^2-1)$

and conclude that $x=3$

Or:

$3^3+4^3+5^3=6^3$