Find the solution of the exponential equation $e^{2 x + 1} = 31$
in terms of logarithms, or correct to four decimal places.
$x =$
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Final Answer: The solution to the equation $e^{2 x + 1} = 31$ is $x = 1.2169936022425731$ , correct to four decimal places.

Steps

Step 1 ：The given equation is $e^{2 x + 1} = 31$ .

Step 2 ：Take the natural logarithm (ln) on both sides of the equation to bring down the exponent on the left side: $l n (e^{2 x + 1}) = l n (31)$ .

Step 3 ：This simplifies to $2 x + 1 = l n (31)$ .

Step 4 ：Isolate x by subtracting 1 and dividing by 2: $x = \frac{l n (31) - 1}{2}$ .

Step 5 ：Using a calculator, we find that $x = 1.2169936022425731$ .

Step 6 ：Final Answer: The solution to the equation $e^{2 x + 1} = 31$ is $x = 1.2169936022425731$ , correct to four decimal places.