Question

$6 {}^{x} + 6 {}^{x + 1} = 2 {}^{x} + 2 {}^{x + 1} + 2 {}^{x + 2}$

Answer 1

[tex]6^x+6^{x+1}=6^x+6^x\cdot6=6^x(1+6)=6^x\cdo\\ 2^x+2^{x+1}+2^{x+2}=2^x+2^{x}\cdot1+2^{x}\cdot2=2^x(1+2+2^2)=2^x\cdot7\\ 6^x+6^{x+1}=2^x+2^{x+1}+2^{x+2}\\ 6^x\cdot7=2^x\cdot7\:|\cdot\frac{1}{7}\\ 6^x=2^x\\ cum\:2^x \ \textgreater \ 0,\:\forall\:x\in\mathbb{R},\:putem\:inmulti\:ecuatia\:cu\:\frac{1}{2^x}\\ \frac{6^x}{2^x}=1\\ \\ \frac{(2\cdot3)^x}{2^x}=1\\ \\ \frac{2^x\cdot3^x}{2^x}=1\\ simplificam\:prin\:2^x\\ 3^x=1\\ logaritmam\:cu\:\log_3\\ \log_3{3^x}=log_3{1}\\ x=0[/tex]