2 ^ { 2 x } - 6.2 ^ { x + 1 } + 32 = 0
Answers
textbf{Given:}Given:
2^{2x}-6.2^{x+1}+32=02
2x
−6.2
x+1
+32=0
\textbf{To find:}To find:
\text{The value of x}The value of x
\textbf{Solution:}Solution:
\text{Consider,}Consider,
2^{2x}-6.2^{x+1}+32=02
2x
−6.2
x+1
+32=0
(2^x)^2-6.2^{x}2^1+32=0(2
x
)
2
−6.2
x
2
1
+32=0
(2^x)^2-12(2^x)+32=0(2
x
)
2
−12(2
x
)+32=0
t^2-12t+32=-0\;\;\text{where}\;t=2^xt
2
−12t+32=−0wheret=2
x
t^2-4t-8t+32=0t
2
−4t−8t+32=0
t(t-4)-8(t-4)=0t(t−4)−8(t−4)=0
(t-8)(t-4)=0(t−8)(t−4)=0
\implies\,t=4,8⟹t=4,8
\begin{gathered}\begin{array}{|c|c|}\cline{1-2}\textbf{Case(i)}&\textbf{Case(ii)}\\\cline{1-2}&\\t=4&t=8\\2^x=4&2^x=8\\2^x=2^2&2^x=2^3\\\implies\bf\,x=2&\implies\bf\,x=3\\\cline{1-2}\end{array}\end{gathered}
\cline1−2Case(i)
\cline1−2
t=4
2
x
=4
2
x
=2
2
⟹x=2
\cline1−2
Case(ii)
t=8
2
x
=8
2
x
=2
3
⟹x=3