Question

any one can solve this ...​

Answer 1

Question :-

$\sf solve \: log_{x}2 = - 1$

Solution :-

$\sf log_{x}2 = - 1$

Write it in exponential form

$\sf \implies x^{ - 1} = 2$

$\boxed{ \bf \because If \: log_{a}N = x \: then \: a^x =N}$

$\sf \implies \dfrac{1}{x^1} = 2$

$\boxed{ \bf \because a^{ - n} = \dfrac{1}{a^n } }$

$\sf \implies \dfrac{1}{x} = 2$

Reciprocal on both sides

$\sf \implies \dfrac{x}{1} = \dfrac{1}{2} \\ \\ \\ \sf \implies x = \dfrac{1}{2} \\ \\ \\ \bf \therefore x = \dfrac{1}{2}$

Verification :-

$\sf log_{x}2 = - 1 \\ \\ \\ \sf \implies log_{ \frac{1}{2} }2 = - 1 \\ \\ \\ \sf \implies \left( \dfrac{1}{2} \right)^{ - 1} = 2 \\ \\ \\ \sf \implies 2^1 = 2 \\ \\ \\ \sf \implies 2 = 2$

Laws of exponents used :-

$\tt \longrightarrow a^{ - n} = \dfrac{1}{a^n }$