17Calculus Precalculus - Solve Exponentials
17Calculus
Solving Equations Involving Exponentials
Recommended Books on Amazon (affiliate links) | ||
|---|---|---|
 |
 |
 |
When we have a variable inside a logarithm, we need to move it out of the logarithm to determine it's value. To do that, we use the laws listed on the laws of logarithms page. Here is how it works.
When we have an exponential, say \(e^x\), to get the variable out of the exponent we take the natural log to get \(\ln (e^x) \). Then we use the law \(\ln(x^y) = y\ln x\). This gives us \(\ln (e^x) = x\ln e \). But \(\ln e = 1\), so we end up with \(\ln (e^x) = x\ln e = x \). Let's go through an example to see this in action.
Example
Solve \(2^x = 30\).
Solution |
\(2^x = 30\) |
Take the natural log of both sides. |
\( \ln(2^x) = \ln(30) \) |
Use the law \(\ln(x^y) = y\ln x\) to simplify the left side. |
\( x\ln(2) = \ln(30) \) |
Solve for \(x\). |
\( x = \ln(30)/\ln(2) \) |
Practice
Unless otherwise instructed, solve these problems using the natural logarithm giving your answers in exact terms.
\( 3^x = 5 \)
Problem Statement
Solve \( 3^x = 5 \)
Solution
3035 video solution
Log in to rate this practice problem and to see it's current rating. |
|---|
\( 5^{2x+3} = 8 \)
Problem Statement
Solve \( 5^{2x+3} = 8 \)
Solution
3036 video solution
Log in to rate this practice problem and to see it's current rating. |
|---|
\( 3^{x+2} = 4^{2-x} \)
Problem Statement
Solve \( 3^{x+2} = 4^{2-x} \)
Solution
3037 video solution
Log in to rate this practice problem and to see it's current rating. |
|---|
\(8^x=15\)
Problem Statement
Solve \(8^x=15\) using the natural logarithm giving your answer in exact terms.
Solution
PatrickJMT - 1698 video solution
video by PatrickJMT |
|---|
Log in to rate this practice problem and to see it's current rating. |
|---|
\(7^x-1=4\)
Problem Statement
Solve \(7^x-1=4\) using the natural logarithm giving your answer in exact terms.
Solution
MIP4U - 1701 video solution
video by MIP4U |
|---|
Log in to rate this practice problem and to see it's current rating. |
|---|
\(3(2^x)-2=13\)
Problem Statement
Solve \(3(2^x)-2=13\) using the natural logarithm giving your answer in exact terms.
Solution
MIP4U - 1702 video solution
video by MIP4U |
|---|
Log in to rate this practice problem and to see it's current rating. |
|---|
\((2/3)^x=5^{3-x}\)
Problem Statement
Solve \((2/3)^x=5^{3-x}\) using the natural logarithm giving your answer in exact terms.
Solution
MIP4U - 1703 video solution
video by MIP4U |
|---|
Log in to rate this practice problem and to see it's current rating. |
|---|
\(5^{x-3}=3^{2x+1}\)
Problem Statement
Solve \(5^{x-3}=3^{2x+1}\) using the natural logarithm giving your answer in exact terms.
Solution
MIP4U - 1704 video solution
video by MIP4U |
|---|
Log in to rate this practice problem and to see it's current rating. |
|---|
\(1111=5(2^t)\)
Problem Statement
Solve \(1111=5(2^t)\) using the natural logarithm giving your answer in exact terms.
Solution
Khan Academy - 1700 video solution
video by Khan Academy |
|---|
Log in to rate this practice problem and to see it's current rating. |
|---|
\(\displaystyle{\left( \frac{4}{5} \right)^x = 6^{1-x}}\)
Problem Statement
Solve \(\displaystyle{\left( \frac{4}{5} \right)^x = 6^{1-x}}\) using the natural logarithm giving your answer in exact terms.
Solution
PatrickJMT - 1699 video solution
video by PatrickJMT |
|---|
Log in to rate this practice problem and to see it's current rating. |
|---|
The exponential function \( V(x) = 25 (4/5)^x \) is used to model the value of a car over time. Use the properties of logarithms and exponentials to rewrite the model in the form \( V(t) = 25e^{kt} \).
Problem Statement |
|---|
The exponential function \( V(x) = 25 (4/5)^x \) is used to model the value of a car over time. Use the properties of logarithms and exponentials to rewrite the model in the form \( V(t) = 25e^{kt} \).
Final Answer |
|---|
\(V(t) = 25e^{t\ln(4/5)}\)
Problem Statement
The exponential function \( V(x) = 25 (4/5)^x \) is used to model the value of a car over time. Use the properties of logarithms and exponentials to rewrite the model in the form \( V(t) = 25e^{kt} \).
Solution
Notice that the two equations are the same except for the exponential terms. So we need to determine \(k\) from \((4/5)^x = e^{kt}\). Although not stated in the problem, since the function names are both V, we can assume that \(x=t\).
\( (4/5)^x = e^{kt} \) |
\( \ln[(4/5)^x] = \ln[e^{kt}] \) |
\( x\ln[(4/5)] = kt\ln[e] \) |
\( x\ln[(4/5)] = kt \) |
\( x\ln[(4/5)] = kt \) |
\( (x/t)\ln[(4/5)] = k \) |
Since \(x=t\), \(x/t = 1\) |
\( \ln[(4/5)] = k \) |
Final Answer
\(V(t) = 25e^{t\ln(4/5)}\)
Log in to rate this practice problem and to see it's current rating. |
|---|
Really UNDERSTAND Precalculus
Log in to rate this page and to see it's current rating.
Topics You Need To Understand For This Page |
|---|
To bookmark this page and practice problems, log in to your account or set up a free account.
Search Practice Problems
Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
|
| |
Join Amazon Prime - Watch Thousands of Movies & TV Shows Anytime - Start Free Trial Now |
|---|
I recently started a Patreon account to help defray the expenses associated with this site. To keep this site free, please consider supporting me. |
|---|
| Support 17Calculus on Patreon |
|
|---|
