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Chapter 3: Problem 31

Solve the exponential equation algebraically. Approximate the result to three decimal places. $$7-2 e^{x}=5$$

Short Answer

Expert verified
The solution to the equation \(7-2 e^{x}=5\) is \(x=0\).

Step by step solution

01

Isolate the exponential term

By subtracting 7 from both sides of the equation \(7-2 e^{x}=5\), you get \(-2 e^{x}=-2\)
02

Remove the coefficient

Then divide by -2 from both sides of the equation to isolate the exponential term. So it gives you the equation \(e^{x}=1\).
03

Take the natural logarithm of both sides

Since the equation is an exponential form, take natural logarithm of both sides, which helps you to remove the exponent. This gives \(\ln(e^{x})=\ln(1)\). Using the logarithmic property \(\ln(a^b) = b \cdot \ln(a)\), this simplifies to \(x\cdot \ln(e)=0\). Since \(\ln(e) = 1\), we have the equation \(x=0\).
04

Approximate to three decimal places

Rounding off some calculations to three decimal places is a necessary skill for many math problems. However, as the solution here is 0, there are no decimals to round off. So, the answer remains 0.

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