Question

Find x,if
\(4^{2x}=\dfrac{1}{32}\)

Answer 1

\(\text{find the value of x}\)
Given \(4^{2x}=\dfrac{1}{32}\)
\(\text{It can be simplified as}\)
\(\left( 2*2 \right)^{2x}=\left( 32 \right)^{-1}\) \(\left[\because \frac{1}{a^m} =a^{-m}\right]\)

\(\left( 2*2 \right)^{2x}=\left( 2*2*2*2*2 \right)^{-1}\)
\(\left(( 2 \right)^2)^{2x}=\left( 2^5 \right)^{-1}\)
\(2^{4x}=2^{-5}\) \(\left[ \left( a^m \right)^n=a^{m*n} \right]\)

We know that:
\(a^{m}=a^{n}\Rightarrow m=n\)
\(\therefore 4x=-5\)
\(\therefore x=-\dfrac{5}{4}\)
Hence, the value of x is \( -\dfrac{5}{4}\).