Solving Equations Involving Exponentials
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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
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\( 5^{2x+3} = 8 \)
Problem Statement
Solve \( 5^{2x+3} = 8 \)
Solution
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\( 3^{x+2} = 4^{2-x} \)
Problem Statement
Solve \( 3^{x+2} = 4^{2-x} \)
Solution
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\(8^x=15\)
Problem Statement
Solve \(8^x=15\) using the natural logarithm giving your answer in exact terms.
Solution
video by PatrickJMT |
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\(7^x-1=4\)
Problem Statement
Solve \(7^x-1=4\) using the natural logarithm giving your answer in exact terms.
Solution
video by MIP4U |
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\(3(2^x)-2=13\)
Problem Statement
Solve \(3(2^x)-2=13\) using the natural logarithm giving your answer in exact terms.
Solution
video by MIP4U |
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\((2/3)^x=5^{3-x}\)
Problem Statement
Solve \((2/3)^x=5^{3-x}\) using the natural logarithm giving your answer in exact terms.
Solution
video by MIP4U |
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\(5^{x-3}=3^{2x+1}\)
Problem Statement
Solve \(5^{x-3}=3^{2x+1}\) using the natural logarithm giving your answer in exact terms.
Solution
video by MIP4U |
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\(1111=5(2^t)\)
Problem Statement
Solve \(1111=5(2^t)\) using the natural logarithm giving your answer in exact terms.
Solution
video by Khan Academy |
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\(\displaystyle{\left( \frac{4}{5} \right)^x = 6^{1-x}}\)
Problem Statement
Solve \(\displaystyle{\left( \frac{4}{5} \right)^x = 6^{1-x}}\) using the natural logarithm giving your answer in exact terms.
Solution
video by PatrickJMT |
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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 |
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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 |
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\(V(t) = 25e^{t\ln(4/5)}\)
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
\(V(t) = 25e^{t\ln(4/5)}\)
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