High School

Solve each equation. Round answers to the nearest ten-thousandth.

[tex]2.75 e^x = 38.6[/tex]

Answer :

We start with the given equation:
[tex]$$2.75 e^x = 38.6.$$[/tex]

Step 1. Isolate [tex]$e^x$[/tex].

Divide both sides of the equation by [tex]$2.75$[/tex]:
[tex]$$e^x = \frac{38.6}{2.75} \approx 14.0364.$$[/tex]

Step 2. Solve for [tex]$x$[/tex].

Take the natural logarithm of both sides:
[tex]$$\ln(e^x) = \ln\left(\frac{38.6}{2.75}\right).$$[/tex]

Since [tex]$\ln(e^x) = x$[/tex], we have:
[tex]$$x = \ln(14.0364).$$[/tex]

Step 3. Evaluate the Natural Logarithm.

Evaluating the logarithm gives:
[tex]$$x \approx 2.6417.$$[/tex]

Thus, the solution to the equation, rounded to the nearest ten-thousandth, is:
[tex]$$x \approx 2.6417.$$[/tex]