### Algebra II
#### 3.13.3 Test (CST): Exponents, Logarithms, & Their Graphs

Question 12 of 20

What is the solution to the equation below? Round your answer to two decimal places.

[tex]\[ e^{0.3x} = 0.3 \][/tex]

A. [tex]\( x = -3.65 \)[/tex]

B. [tex]\( x = -4.01 \)[/tex]

C. [tex]\( x = -1.20 \)[/tex]

D. [tex]\( x = -0.36 \)[/tex]



Answer :

To solve the equation [tex]\(e^{0.3x} = 0.3\)[/tex], we need to isolate [tex]\(x\)[/tex]. Here’s the step-by-step solution:

1. Rewrite the equation in logarithmic form to isolate the exponent:
[tex]\[ e^{0.3x} = 0.3 \][/tex]
Taking the natural logarithm (ln) on both sides gives:
[tex]\[ \ln(e^{0.3x}) = \ln(0.3) \][/tex]

2. Simplify the left side:
Note that [tex]\(\ln(e^y) = y\)[/tex], so:
[tex]\[ 0.3x = \ln(0.3) \][/tex]

3. Solve for [tex]\(x\)[/tex]:
To isolate [tex]\(x\)[/tex], divide both sides by 0.3:
[tex]\[ x = \frac{\ln(0.3)}{0.3} \][/tex]

4. Calculate [tex]\(\ln(0.3)\)[/tex]:
The natural logarithm of 0.3 is approximately [tex]\(-1.2039728043259361\)[/tex].

5. Divide by 0.3:
[tex]\[ x = \frac{-1.2039728043259361}{0.3} \approx -4.013242681086454 \][/tex]

6. Round to two decimal places:
[tex]\[ x \approx -4.01 \][/tex]

So, the solution to the equation [tex]\(e^{0.3x} = 0.3\)[/tex], rounded to two decimal places, is:
[tex]\[ \boxed{-4.01} \][/tex]

Therefore, the correct answer is:
B. [tex]\(x = -4.01\)[/tex]