Question

Solve the following equations.
$(\sqrt[3]{3x}{)}^{lo{g}_x^{11-2}}=1.$

Answer 1

${\left(\sqrt[3]{3x}\right)}^{{log}_x^{11}-2}=1$

$x^{\frac{1}{3}({log}_x^{11}-2)}=1$

$x^{{log}_x^{{11}^{1/3}}-\frac{2}{3}{log}_x^x}=1$

$x^{{log}_x^{{11}^{1/3}}-{log}_x^{x^{2/3}}}=1$

$x^{{log}_x^{\frac{{11}^{1/3}}{x^{2/3}}}}=1$ [As $a^{{log}_a^x}=x$]

$\frac{{11}^{1/3}}{x^{2/3}}=1$

$x^{2/3}={11}^{1/3}$

$x=\sqrt{11}$ (Answer)