Question

Help me solve please 3^1-2x=4^x

Answer 1

$\bf log_{{ a}}(xy)\implies log_{{ a}}(x)+log_{{ a}}(y) \\ \quad \\ % Logarithm of exponentials log_{{ a}}\left( x^{{ b}} \right)\implies {{ b}}\cdot log_{{ a}}(x) \\\\ -----------------------------$

$\bf 3^{1-2x}=4^x\implies 3\cdot 3^{-2x}=4^x\implies 3\cdot \cfrac{1}{(3^2)^x}=4^x \\\\\\ \cfrac{3}{9^x}=4^x\implies 3=9^x\cdot 4^x\impliedby \textit{now taking logarithm} \\\\\\ log(3)=log(9^x\cdot 4^x)\implies log(3)=log(9^x)+log(4^x) \\\\\\ log(3)=xlog(9)+xlog(4)\impliedby \textit{common factor} \\\\\\ log(3)=x[log(9)+log(4)]\implies \boxed{\cfrac{log(3)}{log(9)+log(4)}=x}$