Exponential and Logarithmic Equations
Exponential Equations
Exponential equations are equations that involve a variable in an exponent. The crucial step in solving exponential equations is generally to take the logarithm of both sides to an appropriate base, commonly base 10 or base e.
EXAMPLE 19.1 Solve $ e^{x} = 2 $ .
Take logarithms of both sides
$ (e^{x})=(2) $ Apply the function-inverse function relation
Logarithmic Equations
Logarithmic equations are equations that involve the logarithm of a variable or variable expression. The crucial step in solving logarithmic equations is generally to rewrite the logarithmic statement in exponential form. If more than one logarithmic expression is present, these can be combined into one by using properties of logarithms.
EXAMPLE 19.2 Solve $ _{2}(x-3)=4 $
\[ \begin{aligned}\log_{2}\left(x-3\right)&=4\\2^{4}&=x-3\\x&=2^{4}+3\\x&=19\end{aligned} \]
Rewrite in exponential form
Isolate the variable
Change-of-Base Formula
Logarithmic expressions can be rewritten in terms of other bases by means of the change-of-base formula:
\[ \log_{a}x=\frac{\log_{b}x}{\log_{b}a} \]
EXAMPLE 19.3 Find an expression, in terms of logarithms to base e, for $ _{5}10 $ , and give an approximate value for the quantity.
From the change-of-base formula, $ _{5}10= $
Logarithmic Scales
Working with numbers that range over very wide scales, for example, from 0.000 000 000 001 to 10,000,000,000, can be very cumbersome. The work can be done more efficiently by working with the logarithms of the numbers (as in this example, where the common logarithms range only from -12 to +10).
Examples of Logarithmic Scales
- SOUND INTENSITY: The decibel scale for measuring sound intensity is defined as follows:
\[ D=10\log\frac{I}{I_{0}} \]
where D is the decibel level of the sound, I is the intensity of the sound (measured in watts per square meter), and $ I_{0} $ is the intensity of the smallest audible sound.
- EARTHQUAKE INTENSITY: There is more than one logarithmic scale, called a Richter scale, used to measure the destructive power of an earthquake. A commonly used Richter scale is defined as follows:
\[ R=\frac{2}{3}\log\frac{E}{E_{0}} \]
where R is called the (Richter) magnitude of the earthquake, E is the energy released by the earthquake (measured in joules), and $ E_{0} $ is the energy released by a very small reference earthquake.
SOLVED PROBLEMS
19.1. Prove the change-of-base formula.
Let $ y = _{a} x $ . Then, rewritten in exponential form, $ x = a^{y} $ . Taking logarithms of both sides to the base b yields:
\[ \begin{aligned}\log_{b}x&=\log_{b}a^{y}\\&=y\log_{b}a\quad by the properties of logarithms\end{aligned} \]
Hence,
\[ y=\frac{\log_{b}x}{\log_{b}a},that is,\log_{a}x=\frac{\log_{b}x}{\log_{b}a} \]
19.2. Solve $ 2^{x} = 6 $ .
Take logarithms of both sides to base e (base 10 could equally well be used, but base e is standard in most calculus situations).
\[ \begin{aligned}\ln2^{x}&=\ln6\\x\ln2&=\ln6\\x&=\frac{\ln6}{\ln2}\quad&Exact Answer\\x&\approx2.58\quad&Approximate answer\end{aligned} \]
Alternatively, take logarithms of both sides to base 2 and apply the change-of-base formula:
\[ \begin{aligned}\log_{2}2^{x}&=\log_{2}6\\x&=\log_{2}6\\x&=\frac{\ln6}{\ln2}\end{aligned}\quad by the change-of-base formula \]
19.3. Solve $ 2^{3x-4}=15 $
Proceed as in the previous problem.
\[ \begin{aligned}\ln2^{3x-4}&=\ln15\ $ 3x-4)\ln2&=\ln15\\3x\ln2-4\ln2&=\ln15\\3x\ln2&=\ln15+4\ln2\\x&=\frac{\ln15+4\ln2}{3\ln2}\quad Exact answer\\x&\approx2.64\quad Approximate answer\end{aligned} \]
19.4. Solve $ 5^{4-x} = 7^{3x+1} $
Proceed as in the previous problem.
\[ \begin{aligned}\ln5^{4-x}&=\ln7^{3x+1}\ $ 4-x)\ln5&=(3x+1)\ln7\\4\ln5-x\ln5&=3x\ln7+\ln7\\4\ln5-\ln7&=x\ln5+3x\ln7\\x&=\frac{4\ln5-\ln7}{\ln5+3\ln7}\quad Exact answer\\x&\approx0.60\quad Approximate answer\end{aligned} \]
19.5. Solve $ 2^{x} - 2^{-x} = 1 $ .
Before taking logarithms of both sides it is crucial to isolate the exponential form:
\[ 2^{x}-\frac{1}{2^{x}}=1 \]
Multiply both sides by $ 2^{x} $
\[ 2^{x}\cdot2^{x}-2^{x}\cdot\frac{1}{2^{x}}=2^{x} \]
\[ \begin{aligned}(2^{x})^{2}-1&=2^{x}\ $ 2^{x})^{2}-2^{x}-1&=0\end{aligned} \]
This equation is in quadratic form. Introduce the substitution \(u = 2^{x}\). Then \(u^{2} = (2^{x})^{2}\) and the equation becomes:
\[ u^{2}-u-1=0 \]
Now apply the quadratic formula with a = 1, b = -1, c = -1.
\[ \begin{aligned}u&=\frac{-(-1)\pm\sqrt{(-1)^{2}-4(1)(-1)}}{2(1)}\\&=\frac{1\pm\sqrt{5}}{2}\end{aligned} \]
Now undo the substitution $ 2^{x} = u $ and take logarithms of both sides.
\[ \begin{aligned}2^{x}&=\frac{1\pm\sqrt{5}}{2}\\x\ln2&=\ln\frac{1\pm\sqrt{5}}{2}\\x&=\ln\left(\frac{1+\sqrt{5}}{2}\right)/\ln2\quad&or\quad&\ln\left(\frac{1-\sqrt{5}}{2}\right)/\ln2\end{aligned} \]
Note that since \(\frac{1-\sqrt{5}}{2}\) is negative, it is not in the domain of the logarithm function. Hence the only solution is \(x=\ln\left(\frac{1+\sqrt{5}}{2}\right)/\ln2\) or, approximately, 0.69.
19.6. Solve $ =y $ for x in terms of y.
First note that the left side is a complex fraction (since $ e^{-x} = 1/e^{x} $ ) and write it as a simple fraction.
\[ \frac{e^{x}-1/e^{x}}{e^{x}+1/e^{x}}=y \]
\[ \frac{e^{x}(e^{x}-1/e^{x})}{e^{x}(e^{x}+1/e^{x})}=y \]
\[ \frac{(e^{x})^{2}-1}{(e^{x})^{2}+1}=y \]
\[ \frac{e^{2x}-1}{e^{2x}+1}=y \]
Now, isolate the exponential form $ e^{2x} $ .
\[ e^{2x}-1=y(e^{2x}+1) \]
\[ e^{2x}-1=e^{2x}y+y \]
\[ e^{2x}-e^{2x}y=1+y \]
\[ e^{2x}(1-y)=1+y \]
\[ e^{2x}=\frac{1+y}{1-y} \]
Taking logarithms of both sides yields:
\[ \ln e^{2x}=\ln\frac{1+y}{1-y} \]
\[ 2x=\ln\frac{1+y}{1-y} \]
\[ x=\frac{1}{2}\ln\frac{1+y}{1-y} \]
This is valid as long as the expression $ $ is positive, that is, for -1 < y < 1.
19.7. Solve $ _{2}(3x-4)=5 $
Rewrite the logarithm statement in exponential form, then isolate the variable.
\[ \begin{aligned}2^{5}&=3x-4\\32&=3x-4\\x&=12\end{aligned} \]
19.8. Solve $ x + (x + 3) = 1 $
Use the properties of logarithms to combine the logarithmic expressions into one expression, then rewrite the logarithm statement in exponential form.
\[ \begin{aligned}\log\left[x\left(x+3\right)\right]&=1\\10^{1}&=x\left(x+3\right)\\x^{2}+3x&=10\end{aligned} \]
This quadratic equation is solved by factoring:
\[ \begin{aligned}(x+5)(x-2)&=0\\x=-5or x&=2\end{aligned} \]
Since -5 is not in the domain of the logarithm function, the only solution is 2.
19.9. Solve for y in terms of x and C: $ (y+2)=x+C $
Use the properties of logarithms to combine the logarithmic expressions into one expression, then rewrite the logarithm statement in exponential form.
\[ \begin{aligned}\ln\left(y+2\right)-\ln C&=x\\\ln\left(\frac{y+2}{C}\right)&=x\end{aligned} \]
\[ \frac{y+2}{C}=e^{x} \]
\[ y=Ce^{x}-2 \]
19.10. A certain amount of money P is invested at an annual rate of interest of 4.5%. How many years (to the nearest tenth of a year) would it take for the amount of money to double, assuming interest is compounded quarterly?
Use the formula $ A(t)=P(1+)^{nt} $ from Chapter 17, with n=4 and r=0.045, to find t when $ A(t)=2P $ .
\[ \begin{aligned}2P&=P\bigg(1+\frac{0.045}{4}\bigg)^{4t}\\2&=\bigg(1+\frac{0.045}{4}\bigg)^{4t}\end{aligned} \]
To isolate t, take logarithms of both sides to base e.
\[ \begin{aligned}\ln2&=4t\ln\left(1+\frac{0.045}{4}\right)\\t&=\frac{\ln2}{4\ln\left(1+\frac{0.045}{4}\right)}\\t&\approx15.5years\end{aligned} \]
19.11. In the previous problem, how many years (to the nearest tenth of a year) would it take for the amount of money to double, assuming interest is compounded continuously?
Use the formula $ A(t) = Pe^{rt} $ from Chapter 17, with r = 0.045, to find t when $ A(t) = 2P $ .
\[ \begin{aligned}2P&=Pe^{0.045t}\\2&=e^{0.045t}\end{aligned} \]
To isolate t, take logarithms of both sides to base e.
\[ \begin{aligned}\ln2&=0.045t\\t&=\frac{\ln2}{0.045}\\t&\approx15.4years\end{aligned} \]
19.12. A radioactive isotope has a half-life of 35.2 years. How many years (to the nearest tenth of a year) would it take before an initial quantity of 1 gram decays to 0.01 gram?
Use the formula $ Q(t) = Q_{0}e^{-kt} $ from Chapter 17.
First, determine k by using t = 35.2, $ Q_{0} = 1 $ , and $ Q(35.2) = Q_{0}/2 = 1/2 $ .
\[ 1/2=1e^{-k(35.2)} \]
To isolate k, take logarithms of both sides to base e.
\[ \begin{aligned}\ln\left(1/2\right)&=-k(35.2)\\k&=-\frac{\ln\left(1/2\right)}{35.2}\\&=\frac{\ln2}{35.2}\end{aligned} \]
Thus, for this isotope, the quantity remaining after t years is given by:
\[ Q(t)=Q_{0}e^{\frac{-t\ln2}{35.2}} \]
To find the time required for the initial quantity to decay to 0.01 gram, use this formula with $ Q(t) = 0.01 $ , $ Q_{0} = 1 $ and solve for t.
\[ 0.01=1e^{\frac{-t\ln2}{35.2}} \]
To isolate t, take logarithms of both sides to base e.
\[ \begin{aligned}\ln0.01&=\frac{-t\ln2}{35.2}\\t&=\frac{-35.2\ln0.01}{\ln2}\\t&\approx233.9years\end{aligned} \]
19.13. (a) Calculate the decibel level of the smallest audible sound, $ I_{0} = 10^{-12} $ watts per square meter.
Calculate the decibel level of a rock concert at an intensity of $ 10^{-1} $ watts per square meter.
Calculate the intensity of a sound with decibel level 85.
Use the formula $ D = 10 $
Set $ I = I_{0} $ . Then $ D = 10 = 10 = 0 $ .
Set \(I = 10^{-1}\) and \(I_{0} = 10^{-12}\). Then \(D = 10 \log \frac{10^{-1}}{10^{-12}} = 10 \log 10^{11} = 10 \cdot 11 = 110\) decibels.
Set D = 85 and $ I_{0} = 10^{-12} $ . Then $ 85 = 10 $ . Solving for I yields:
\[ \begin{aligned}8.5&=\log\frac{I}{10^{-12}}\\\frac{I}{10^{-12}}&=10^{8.5}\\I&=10^{-12}\cdot10^{8.5}\\I&=10^{-3.5}\\I&\approx3.2\times10^{-4}watts/square meter\end{aligned} \]
19.14. (a) Find the Richter scale magnitude of an earthquake that releases energy of $ 1000E_{0} $ . (b) Find the energy released by an earthquake that measures 5.0 on the Richter scale, given that $ E_{0} = 10^{4.40} $ joules. (c) What is the ratio in energy released between an earthquake that measures 8.1 on the Richter scale and an aftershock measuring 5.4 on the scale?
Use the formula $ R = $ .
Set $ E = 1000E_{0} $ . Then $ R = = = = 2 $ .
Set R = 5. Then $ 5 = $ . Solving for E yields:
\[ \begin{aligned}\frac{15}{2}&=\log\frac{E}{E_{0}}\\\frac{E}{E_{0}}&=10^{15/2}\\E&=E_{0}\cdot10^{7.5}\\&=10^{4.40}\cdot10^{7.5}\\&\approx7.94\times10^{11}joules\end{aligned} \]
- First, solve the formula for E in terms of R and $ R_{0} $ .
\[ \log\frac{E}{E_{0}}=\frac{3R}{2} \]
\[ \frac{E}{E_{0}}=10^{3R/2} \]
\[ E=E_{0}10^{3R/2} \]
Then set $ R_{1} = 8.1 $ and $ R_{2} = 5.4 $ and find the ratio of the corresponding energies $ E_{1} $ and $ E_{2} $ .
\[ E_{1}=E_{0}10^{3R_{1}/2}\qquad E_{2}=E_{0}10^{3R_{2}/2} \]
\[ E_{1}=E_{0}10^{3(8.1)/2}\qquad E_{2}=E_{0}10^{3(5.4)/2} \]
\[ E_{1}/E_{2}=(E_{0}10^{3(8.1)/2})/(E_{0}10^{3(5.4)/2}) \]
\[ E_{1}/E_{2}=10^{12.15}/10^{8.1} \]
\[ E_{1}/E_{2}=10^{4.05}/1 \]
\[ E_{1}/E_{2}\approx11,200/1 \]
The energy released by the earthquake is more than 11,000 times the energy released by the aftershock.
SUPPLEMENTARY PROBLEMS
19.15. Show that $ a{b}=e{ba} $
19.16. Solve (a) $ e^{5x-3}=10 $ ; (b) $ 5{3+x}=20{x-3} $ ; (c) $ 4{x{2}-2x}=12 $ .
Ans. (a) $ x = $ ; (b) $ x = $ ;
\[ (c)x=1\pm\sqrt{1+\frac{\ln12}{\ln4}};x\approx2.67,-0.67 \]
19.17. Solve in terms of logarithms to base 10: (a) $ 2^{x} - 6(2^{-x}) = 6 $ ; (b) $ = $ .
Ans. (a) $ x = (3 + ) / $ ; (b) $ x = () / 2 $
19.18. Solve: (a) $ {3}(x-2) + {3}(x-4) = 2 $ ; (b) $ 2 x - (x+1) = 3 $
Ans. (a) $ x = 3 + $ ; (b) $ x = $
19.19. Solve for t, using natural logarithms: (a) Q = Q0ekt; (b) A = P(1 + (r/n)
Ans. (a) $ t = $ ; (b) $ t = $
19.20. Solve for t, using natural logarithms: (a) $ I = ( 1 - e^{-Rt/L} ) $ ; (b) $ N = $
Ans. (a) $ t = $ ; (b) $ t = $
19.21. Solve for x in terms of y: (a) $ = y $ ; (b) $ = y $ .
Ans. (a) $ x = (y ) $ ; (b) $ x = (y + ) $
19.22. How many years would it take an investment to triple at 6% interest compounded quarterly?
Ans. 18.4 years
19.23. At what rate of interest would an investment double in eight years, compounded continuously?
Ans. 8.66%
19.24. If a sample of a radioactive isotope decays from 400 grams to 300 grams in 5.3 days, find the half-life of this isotope.
Ans. 12.8 days
19.25. If the intensity of one sound is 1000 times the intensity of another sound, what is the difference in the decibel level of the two sounds?
Ans. 30 decibels
19.26. Newton’s law of cooling states that the temperature \(T\) of a body, initially at temperature \(T_{0}\), placed in a surrounding medium at a lower temperature \(T_{m}\), is given by the formula \(T = T_{m} + (T_{0} - T_{m})e^{-kt}\). If a cup of coffee, at temperature \(160^{\circ}\) at \(7\;\mathrm{a.m.}\), is brought outside into air at \(40^{\circ}\), and cools to \(140^{\circ}\) by \(7:05\;\mathrm{a.m.}\), (a) find its temperature at \(7:10\;\mathrm{a.m.}\) (b) At what time will the temperature have fallen to \(100^{\circ}\)?
Ans. (a) \(123^{\circ}\); (b) \(7:19\;\mathrm{a.m.}\)