[tex]\text{Am gasit o solutie mai smechera:}\\
6^x+8^x+15^x=9^x+12^x+10^x\\
(6^x-9^x)+(8^x-12^x)+(15^x-10^x)=0\\
3^x(2^x-3^x)+4^x(2^x-3^x)-5^x(2^x-3^x)=0\\
(2^x-3^x)(3^x+4^x-5^x)=0\\
\text{Egalam fiecare paranteza cu 0:}\\
2^x-3^x=0\\
2^x=3^x\Rightarrow x=0\\
3^x+4^x-5^x=0\\
3^x+4^x=5^x |:5^x\neq 0\\
\left(\dfrac{3}{5}\right)^x+\left(\dfrac{4}{5}\right)^x=1\\
[/tex]
[tex]\text{Consideram functia:} f:\mathbb{R}\rightarrow (0,\infty),f(x)=\left(\dfrac{3}{5}\right)^x+\left(\dfrac{4}{5}\right)^x\\
\text{Deoarece f este injectiva inseamna ca ecuatia f(x)=1 are o singura}\\ \text{solutie.Se observa ca f(2)=1,deci x=2 este singura solutie.}\\
S:x\in \{0,2\}[/tex]
