Approximate the solution to the equation 20=125e^x

Question

by locating the intersection point of y=125e^x
and y=20
.(1 point)
Responses

x≈20
x approximately equals 20

x≈−1.8
x approximately equals negative 1.8

x≈125
x approximately equals 125

x≈1.8
x approximately equals 1.8

Answer 1

To solve the equation \( 20 = 125e^x \), we can rewrite it as:

\[ e^x = \frac{20}{125} \]

Simplifying the fraction:

\[ e^x = \frac{4}{25} \]

Now, take the natural logarithm of both sides:

\[ x = \ln\left(\frac{4}{25}\right) \]

Using properties of logarithms:

\[ x = \ln(4) - \ln(25) \]

We can approximate this value. We know that:

\[ \ln(4) \approx 1.386 \quad \text{and} \quad \ln(25) = \ln(5^2) = 2\ln(5) \approx 2 \times 1.609 = 3.218 \]

So:

\[ x \approx 1.386 - 3.218 \approx -1.832 \]

This indicates that \( x \) is approximately \(-1.8\).

Thus, the correct response is:

x ≈ -1.8.