Question

Solve the following equation:

$1-\log 5=\frac{1}{3}(\log \frac{1}{2}+\log x+\frac{1}{3}\log 5)$

Answer 1

$1-log5=\frac{1}{3}(log\frac{1}{2}+logx+\frac{1}{3}log5)$

$\Rightarrow log10-log5=\frac{1}{3}[-log2+logx+\frac{1}{3}(1-log2\left)\right]$

$\Rightarrow log2=\frac{1}{3}(-log2+logx+\frac{1-log2}{3})$

$\Rightarrow 3log2=-log2+logx+\frac{1}{3}-\frac{log2}{3}$

$\Rightarrow 9log2=-3log2+3logx+1-log2$

$\Rightarrow 13log2-1=3logx$

$\Rightarrow x^3=\frac{2^{12}}{10}$

$\Rightarrow x=\frac{16}{\sqrt[3]{35}}$

$\Rightarrow x=16\cdot 5^{\frac{-1}{3}}$