Question

Solve the equations
(i) ${\log }_{5}x=-3$
(ii) $x={\log }_{\frac{1}{4}}64$
(iii) ${\log }_{x}8=2$
(iv) ${\log }_x{7}^{\frac{1}{6}}=\frac{1}{3}$

Answer 1

(i) ${\log }_{5}x=-3$
$5^{-3}=x$ (exponential form)
$x=\frac{1}{5^3}⟹x=\frac{1}{125}$
(ii) $x={\log }_{\frac{1}{4}}64$
${\left(\frac{1}{4}\right)}^x=64$ (exponential form)
$\frac{1}{4^x}=4^3⟹4^{-x}=4^3⟹x=-3$
(iii) ${\log }_{x}8=2$
$x^2=8$ (exponential form)
$x=\sqrt{8}=2\sqrt{2}$
(iv) ${\log }_x{7}^{\frac{1}{6}}=\frac{1}{3}⟹x^{\frac{1}{3}}=7^{\frac{1}{6}}$ (exponential form)
We write $7^{\frac{1}{6}}={\left(7^{\frac{1}{2}}\right)}^{\frac{1}{3}}$. Then $x^{\frac{1}{3}}={\left(7^{\frac{1}{2}}\right)}^{\frac{1}{3}}$
$\therefore x=7^{\frac{1}{2}}=\sqrt{7}$