Solving Exponential Equations
Back in Target C, we talked about the difference between an equation like \(5^x=125\) and \(5^x=95\). Both are exponential equations where is a variable in the exponent. The first equation is solved by thinking of like bases (or \(5^x=5^3\) ) so we know \(x=3\). Now that you know about logarithms, you will be able to solve the equation \(5^x=95\).
Let’s first look at some examples of exponential equations that can be re-written using like bases.
Example 1:
\(9^{x-5}=27^x\;\;\;\;\;\;\;\; \text{ Both \(9\) and \(27\) can be rewritten using a base of } 3.\\\\
3^{2(x-5)}=3^{3x}\;\;\;\;\;\; \text{Be careful to multiply the entire original exponent with the new exponent.}\\\\
2(x-5)=3x\;\;\;\; \text{Since we now have like bases and know these expressions are equivalent,}\\\\
\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{then the exponents must be equivalent as well.}\\\\
2x-10=3x\\\\
x=-10\)
Example 2: Solve \(4^{3x+5}=\left(\frac{1}{8}\right)^{2x}\).
Back in Target C, we talked about the difference between an equation like \(5^x=125\) and \(5^x=95\). Both are exponential equations where is a variable in the exponent. The first equation is solved by thinking of like bases (or \(5^x=5^3\) ) so we know \(x=3\). Now that you know about logarithms, you will be able to solve the equation \(5^x=95\).
Let’s first look at some examples of exponential equations that can be re-written using like bases.
Example 1:
\(9^{x-5}=27^x\;\;\;\;\;\;\;\; \text{ Both \(9\) and \(27\) can be rewritten using a base of } 3.\\\\
3^{2(x-5)}=3^{3x}\;\;\;\;\;\; \text{Be careful to multiply the entire original exponent with the new exponent.}\\\\
2(x-5)=3x\;\;\;\; \text{Since we now have like bases and know these expressions are equivalent,}\\\\
\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{then the exponents must be equivalent as well.}\\\\
2x-10=3x\\\\
x=-10\)
Example 2: Solve \(4^{3x+5}=\left(\frac{1}{8}\right)^{2x}\).
Let’s now go back to the equation \(5^x=95\). Since we can’t rewrite using the same base, we have to take a different approach. If we take the log of each side, we get \(\log {5^x}=\log{95}\) and we can use the power property of logs to rewrite as \(x\log{5}=\log{95}\). So the exact solution is \(x=\frac{\log{95}}{\log{5}}\) and we can use our calculator to get an approximate solution of \(x=2.829\). It would also work to take the natural log of both sides or another log with a different base, but we will mostly use common logs (base \(10\)) or natural logs (base \(e\)) due to the ease of typing those into the calculator.
Example 3:
\(\frac{1}{3}(7^{2x})=51\\\\
7^{2x}=153\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\; \text{Multiply both sides by 3 to isolate the exponential term.}\\\\
\log{7^{2x}}=\log{153}\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\; \text{Take the log of both sides.}\\\\
2x\log{7}=\log{153}\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\; \text{Use the power property to rewrite by moving the exponent in front.}\\\\
x=\frac{\log{153}}{2\log{7}}\;\; \text{or}\;\; x=1.293\;\;\;\;\;\;\;\;\; \text{Note that}\; x=\frac{\ln{153}}{2\ln{7}}\; \text{would have yielded the same result.}\)
There are a couple special cases to take note of and that is when your exponential equation has a base of \(10\) or a base of \(e\). If the base is \(10\), make sure to take the common log of both sides since \(\log{10}=1\) and if the base is \(e\), make sure to take the natural log of both sides since \(\ln{e}=1\).
Example 4: Solve each equation: a) \(10^{x+5}=7\) b) \(3e^{2x+1}=14\)
We can also now solve some exponential growth and decay application problems using logs instead of graphing.
For your 5th birthday, a generous family member gives you a gift of \(\$5,000\) to put towards college. Your parents have you invest it in an account that earns \(1.53\%\) interest compounded semi-annually. How long will it take before the amount in the account reaches \(\$6,000\)?
\(6000=5000(1+\frac{0.0153}{2})^{2x}\;\;\;\;\;\;\;\;\;\;\; \text{Set-up the equation in the same way discussed in Target B.}\\\\
1.2=(1.00765)^{2x}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ \text{Divide both sides by 5000 to isolate the exponential term and rewrite the expression in}\\\\
\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{parentheses for simplicity.}\\\\
\ln{1.2}=2x\ln{1.00765}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \text{Take the log or natural log of both sides and apply the power property.}\\\\\\\\\\\\
x=\frac{\ln{1.2}}{2\ln{1.00765}}\; \text{or approximately}\;\; 11.962\; \text{or}\; 12 \text{years}\)
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