Provide an appropriate response.
-Solve $2^{2 x + 1 }= 7$ for $x$. Give an exact solution and a four-decimal-place approximation.
A) $\frac { \log 6 } { 2 \log 2 } ; 1.2925$
B) $\frac { 5 } { 4 } ; 1.2500$
C) $\frac { 1 } { 2 } \left( \frac { \log 2 } { \log 7 } - 1 \right) ; - 0.3219$
D) $\frac { 1 } { 2 } \left( \frac { \log 7 } { \log 2 } - 1 \right) ; 0.9037$

Question

Quizplus · Accepted Answer

To solve $2^{2x + 1} = 7$, we first isolate the exponential expression: $2^{2x + 1} = 7$. Taking the logarithm of both sides gives $2x + 1 = \frac{\log 7}{\log 2}$. Solving for $x$ yields $x = \frac{1}{2} \left( \frac{\log 7}{\log 2} - 1 \right)$. The approximation of this value is 0.9037.

Provide an Appropriate Response 22x+1=72^{2 x + 1 }= 722x+1=7 For xxx

Provide an Appropriate Response $2^{2 x + 1 }= 7$ For $x$