If \(2^{2^x}=16^{2^{3x}}\), then \(x=?\)

Given: \(2^{2^x}=16^{2^{3x}}\)

Step 1: Use the concept of prime factorization to write the number as the product of its power:

\(16=2×2×2×2=2^4\)

\(\Rightarrow 16^{2^{3x}}=(2^4)^{2^{3x}}\)

\(\Rightarrow 16^{2^{3x}}=(2^{2^2})^{2^{3x}}\)

Step 2: Use the appropriate laws of exponents to simplify the expressions in R.H.S:

\(16^{2^{3x}}=(2)^{2^2×2^{3x}}\)                   [ \( (a^m)^n=a^{mn} \)]

\(16^{2^{3x}}=(2)^{2^{2+3x}}\)                      [ \(a^m× a^n=a^{m+n}\)]

So, \(2^{2^x}=2^{2^{2+3x}}\)

Let \(2^{2^x}=A\), then

\(2^{2^x}=2^{2^{2+3x}} \Rightarrow A^x=A^{2+3x}\)

\(\Rightarrow x= 2+3x\)                    [\(a^m=a^n \iff m=n\)]

Step 3: Solve the equation for \(x\)

\(x-3x=2\)

\(\Rightarrow -2x=2\)

\(\Rightarrow x=\frac{2}{-2}\)

\(\Rightarrow x=-1\)

Therefore, \(x=-1\).

Verification: For \(x=-1\)

L.H.S = \(2^{2^x}\)= \(2^{2^{-1}}\)= \(2^{\frac{1}{2}}\)= \(\sqrt{2}\)

R.H.S = \(16^{2^{3x}}\)= \(16^{2^{3(-1)}}\)= \(16^{2^{-3}}\)= \(16^{\frac{1}{2^3}}\)

= \(16^{\frac{1}{8}}\)= \(2^{4^{\frac{1}{8}}}\)= \(2^{\frac{1}{2}}\)= \(\sqrt{2}\)

L.H.S=R.H.S=\(\sqrt{2}\)

Therefore, \(2^{2^x}=16^{2^{3x}} \Rightarrow x=-1\)