Question

lesson 28 - exponential and logarithmic equations wi
score: 62/100 answered: 12/20
question 13
find the solution of the exponential equation
$e^{1 - 4x} = 6$
in terms of logarithms, or correct to four decimal places.
$x = $
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Explanation:

Step1: Take natural log of both sides

To solve \( e^{1 - 4x}=6 \), we take the natural logarithm (ln) of both sides. Recall that \( \ln(e^y)=y \) for any real number \( y \). So, applying \( \ln \) to both sides gives:
\( \ln(e^{1 - 4x})=\ln(6) \)
By the property of logarithms, the left - hand side simplifies to \( 1 - 4x \), so we have:
\( 1-4x = \ln(6) \)

Step2: Solve for x

First, we isolate the term with \( x \). Subtract 1 from both sides of the equation \( 1-4x=\ln(6) \):
\( - 4x=\ln(6)-1 \)
Then, divide both sides by - 4 to solve for \( x \):
\( x=\frac{1 - \ln(6)}{4} \)

Now, we calculate the numerical value. We know that \( \ln(6)\approx1.791759 \).
Substitute this value into the formula for \( x \):
\( x=\frac{1 - 1.791759}{4}=\frac{- 0.791759}{4}\approx - 0.1979 \)

Answer:

\( x=\frac{1 - \ln(6)}{4}\approx - 0.1979 \) (or in exact form \( \frac{1-\ln(6)}{4} \))