Question

Solve allgebraically for x: log base 27 (2x-1)=4/3

Answer 1

$log_{27}(2x-1)=\frac{4}{3};\ \ \ \ D:2x-1 > 0\to x > \frac{1}{2}\\\\log_{27}(2x-1)=log_{27}27^\frac{4}{3}\iff 2x-1=27^\frac{4}{3}\\\\2x-1=\left(\sqrt[3]{27}\right)^4\\\\2x-1=3^4\\\\2x-1=81\ \ \ /+1\\\\2x=82\ \ \ \ /:2\\\\x=41\in D$