Chapter10: Exponential And Logarithmic Functions
Section: Chapter Questions
Problem 445RE: Mouse populations can double in 8 months (A=2A0) . How long will it take for a mouse population to...
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Equations and Inequations
Equations and inequalities describe the relationship between two mathematical expressions.
Linear Functions
A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.
Question
![### Problem 1: Solve for \( x \)
\[
35 - 12e^{5 - 4x} = 0
\]
#### Solution Steps:
1. **Isolate the exponential term**:
\[
35 - 12e^{5 - 4x} = 0
\]
Subtract 35 from both sides:
\[
-12e^{5 - 4x} = -35
\]
Divide both sides by -12:
\[
e^{5 - 4x} = \frac{35}{12}
\]
2. **Apply the natural logarithm (ln) to both sides**:
\[
\ln\left(e^{5 - 4x}\right) = \ln\left(\frac{35}{12}\right)
\]
Using the property of logarithms \( \ln(e^y) = y \):
\[
5 - 4x = \ln\left(\frac{35}{12}\right)
\]
3. **Solve for \( x \)**:
Subtract 5 from both sides:
\[
-4x = \ln\left(\frac{35}{12}\right) - 5
\]
Divide both sides by -4:
\[
x = \frac{5 - \ln\left(\frac{35}{12}\right)}{4}
\]
#### Final Answer:
\[
x = \frac{5 - \ln\left(\frac{35}{12}\right)}{4}
\]
This equation demonstrates solving for \( x \) in an exponential equation by isolating the exponential term, applying logarithms, and then solving the resulting linear equation.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb3c4489a-e4bd-4ff1-89b5-d5833f06911e%2Fcfe5a144-e2f1-4132-8823-7eab908e311e%2Fe6pg2dr_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem 1: Solve for \( x \)
\[
35 - 12e^{5 - 4x} = 0
\]
#### Solution Steps:
1. **Isolate the exponential term**:
\[
35 - 12e^{5 - 4x} = 0
\]
Subtract 35 from both sides:
\[
-12e^{5 - 4x} = -35
\]
Divide both sides by -12:
\[
e^{5 - 4x} = \frac{35}{12}
\]
2. **Apply the natural logarithm (ln) to both sides**:
\[
\ln\left(e^{5 - 4x}\right) = \ln\left(\frac{35}{12}\right)
\]
Using the property of logarithms \( \ln(e^y) = y \):
\[
5 - 4x = \ln\left(\frac{35}{12}\right)
\]
3. **Solve for \( x \)**:
Subtract 5 from both sides:
\[
-4x = \ln\left(\frac{35}{12}\right) - 5
\]
Divide both sides by -4:
\[
x = \frac{5 - \ln\left(\frac{35}{12}\right)}{4}
\]
#### Final Answer:
\[
x = \frac{5 - \ln\left(\frac{35}{12}\right)}{4}
\]
This equation demonstrates solving for \( x \) in an exponential equation by isolating the exponential term, applying logarithms, and then solving the resulting linear equation.
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