Problem 33 - College Algebra Essentials

Question

Solve each exponential equation. Express the solution set in terms of natural logarithms or common logarithms. Then use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$e^{1-5 x}=793$$

Step-by-Step Solution

Answer

The solution to the equation is approximately \( x \approx 1.61 \) when rounded to two decimal places.

Step 1: Transforming the equation to a logarithmic form

We start by taking natural logarithms on both sides of the equation to bring down the exponential term, resulting in \( \ln(e^{1-5x}) = \ln(793) \). Since the natural logarithm of \( e \) to any exponent equals that exponent, the equation simplifies to \( 1-5x = \ln(793) \).

Step 2: Solving the equation for 'x'

Next, move the 1 to the other side by subtracting 1 from both sides, the equation becomes: \( -5x = \ln(793) -1 \). Now, to find the value of 'x', we simply divide both sides by -5 and get the value: \( x = \frac{\ln (793) -1}{-5} \).

Step 3: Approximate the value of 'x' to two decimal places

Taking natural log or common log (base 10) of 793 and following the equation to find 'x' value, then rounding 'x' to the nearest hundredths place using a calculator: \( x = \frac{\ln(793) -1 }{-5} \) gives \( x \approx 1.61 \).