Question

Approximate the solution to the equation 20=125e
x
by locating the intersection point of y=125e
x
and y=20.
A. x≈20
B. x≈125
C. x≈−1.8
D. x≈1.8

Answer 1

To find the approximate solution to the equation \( 20 = 125e^x \), we can set up the equations \( y = 125e^x \) and \( y = 20 \) and find their intersection.

1. Start with the equation:
\[
125e^x = 20
\]

2. Divide both sides by 125:
\[
e^x = \frac{20}{125} = \frac{4}{25}
\]

3. Take the natural logarithm of both sides:
\[
x = \ln\left(\frac{4}{25}\right)
\]

4. We can simplify this:
\[
x = \ln(4) - \ln(25)
\]

5. Using approximate values for the logarithms:
- \( \ln(4) \approx 1.386 \)
- \( \ln(25) = \ln(5^2) = 2 \ln(5) \) and \( \ln(5) \approx 1.609 \) hence, \( \ln(25) \approx 2 \times 1.609 = 3.218 \)

So now calculating \( x \):
\[
x \approx 1.386 - 3.218 \approx -1.832
\]

Thus, the approximate solution is:
\[
x \approx -1.8
\]
So, the answer is C. \( x \approx -1.8 \).