Question

Equação exponencial .$\boxed{4*3^{x-5}= \sqrt{2} }$

Answer 1

E aí mano,

$4\cdot3^{x-5}= \sqrt{2}\\ log(4\cdot3^{x-5})=log2^{ \tfrac{1}{2} }\\ log4+(log3^{x-5})= \dfrac{1}{2}\cdot log2\\ log4+[log3\cdot(x-5)]= \dfrac{1}{2}\cdot log2\\\\ \begin{cases}log4=0,602\\ log3=0,477\\ log2=0,301\end{cases}\\\\\\ 0,602+[0,477\cdot(x-5)]= \dfrac{1}{2}\cdot0,301\\ 0,602+0,477x-2,385=0,1505\\ 0,477x-1,783=0,1505\\ 0,477x=1,9335\\\\ x= \dfrac{1,9335}{0,477}\\\\x\approx4,0534\\\\\\ \Large\boxed{\boxed{S\approx\{4,0534\}}}$

FLW 452!