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
\(2^{2^x}=2^{2^{2+3x}} \Rightarrow A^x=A^{2+3x}\)
\(\Rightarrow x= 2+3x\)Â Â Â Â Â Â Â Â Â Â [\(a^m=a^n \iff m=n\)]
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, \(2^{2^x}=16^{2^{3x}} \Rightarrow x=-1\)