0
Maths Expert_1
If \(2^{2^x}=16^{2^{3x}}\), then \(x=?\)
0 Subscribers
Submit Answer
1 Answers
Best Answer
0
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, \(2^{2^x}=16^{2^{3x}} \Rightarrow x=-1\)
Replying as
Submit