Find the solution of the exponential equation $e^{2 x+1}=31$
in terms of logarithms, or correct to four decimal places.
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Final Answer: The solution to the equation \(e^{2 x+1}=31\) is \(x=\boxed{1.2169936022425731}\), correct to four decimal places.
Step 1 ：The given equation is \(e^{2x+1} = 31\).
Step 2 ：Take the natural logarithm (ln) on both sides of the equation to bring down the exponent on the left side: \(ln(e^{2x+1}) = ln(31)\).
Step 3 ：This simplifies to \(2x + 1 = ln(31)\).
Step 4 ：Isolate x by subtracting 1 and dividing by 2: \(x = \frac{ln(31) - 1}{2}\).
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 \(x=\boxed{1.2169936022425731}\), correct to four decimal places.
