Logarithmic Functions
Definition of Logarithmic Function
A logarithmic function, $ f(x) = {a} x $ , $ a > 0 $ , $ a $ , is the inverse function to an exponential function $ F(x) = a^{x} $ . Thus, if $ y = {a} x $ , then $ x = a^{y} $ . That is, the logarithm of x to the base a is the exponent to which a must be raised to obtain x. Conversely, if $ x = a^{y} $ , then $ y = _{a} x $ .
EXAMPLE 18.1 The function $ f(x) = {2} x $ is defined as $ f: y = {2} x $ if $ 2^{y} = x $ . Since $ 2^{4} = 16 $ , 4 is the exponent to which 2 must be raised to obtain 16, and $ _{2} 16 = 4 $ .
EXAMPLE 18.2 The statement $ 10^{3} = 1000 $ can be rewritten in terms of the logarithm to the base 10. Since 3 is the exponent to which 10 must be raised to obtain $ 1000, _{10} 1000 = 3 $ .
Relation between Logarithmic and Exponential Functions
\[ \log_{a}a^{x}=x\quad a^{\log_{a}x}=x \]
EXAMPLE 18.3 $ {5}5{3}=3;5{{5}25}=25 $
Properties of Logarithms
$ (M, N $ positive real numbers $
\[ \log_{a}1=0 \]
\[ \log_{a}(MN)=\log_{a}M+\log_{a}N \]
\[ \log_{a}a=1 \]
\[ \log_{a}\left(M^{p}\right)=p\log_{a}M \]
\[ \log_{a}\left(\frac{M}{N}\right)=\log_{a}M-\log_{a}N \]
EXAMPLE 18.4 (a) $ {5}1=0 $ (since $ 5^{0}=1 $ ) (b) $ {4}4=1 $ (since $ 4^{1}=4 $ )
\[ \begin{aligned}(c)\log_{6}6x&=\log_{6}6+\log_{6}x=1+\log_{6}x\quad(d)\log_{6}x^{6}=6\log_{6}x\end{aligned} \]
\[ \left(\mathrm{e}\right)\log_{1/2}(2x)=\log_{1/2}\frac{x}{1/2}=\log_{1/2}x-\log_{1/2}\left(\frac{1}{2}\right)=\log_{1/2}x-1 \]
Special Logarithmic Functions
$ {10} x $ is abbreviated as $ x $ (common logarithm). $ {e} x $ is abbreviated as $ x $ (natural logarithm).
SOLVED PROBLEMS
18.1. Write the following in exponential form:
- $ {2}8 = 3 $ ; (b) $ {25}5 = $ ; (c) $ _{10} = -2 $ ;
\[ \left(\mathrm{d}\right)\log_{8}\frac{1}{4}=-\frac{2}{3};\left(\mathrm{e}\right)\log_{b}c=d;\left(\mathrm{f}\right)\log_{e}\left(x^{2}+5x-6\right)=y-C \]
If $ y = {a} x $ , then $ x = a^{y} $ . Hence, if $ 3 = {2} 8 $ , then $ 8 = 2^{3} $ .
If $ y = {a} x $ , then $ x = a^{y} $ . Hence, if $ = {25} 5 $ , then $ 5 = 25^{1/2} $ .
If $ -2 = _{10} $ , then $ = 10^{-2} $ .
If $ - = _{8} $ , then $ 8^{-2/3} = $ .
If $ d = _{b} c $ , then $ b^{d} = c $ .
If $ y - C = _{e}(x^{2} + 5x - 6) $ , then $ e^{y - C} = x^{2} + 5x - 6 $ .
18.2. Write the following in logarithmic form:
$ 3^{5} = 243 $ ; (b) $ 6^{-3} = $ ; (c) $ 256^{3/4} = 64 $ ;
$ ()^{-5}=32 $ ; (e) $ u^{m}=p $ ; (f) $ e^{at+b}=y-C $
If $ x = a^{y} $ , then $ y = {a} x $ . Hence if $ 243 = 3^{5} $ , then $ 5 = {3} 243 $ .
If $ x = a^{y} $ , then $ y = {a} x $ . Hence if $ = 6^{-3} $ , then $ -3 = {6} $ .
If $ 64 = 256^{3/4} $ , then $ _{256} 64 = $ .
If $ 32 = ()^{-5} $ , then $ _{1/2} 32 = -5 $ .
If $ p = u^{m} $ , then $ _{u} p = m $ .
If $ y - C = e^{at + b} $ , then $ _{e}(y - C) = at + b $ .
18.3. Evaluate the following logarithms:
$ {7}49 $ ; (b) $ {4}256 $ ; (c) $ {10}0.000001 $ ; (d) $ {27} $ ; (e) $ _{1/5}125 $
The logarithm to base 7 of 49 is the exponent to which 7 must be raised to obtain 49. This exponent is 2; hence \(\log_{7}49=2\).
The logarithm to base 4 of 256 is the exponent to which 4 must be raised to obtain 256. This exponent is 4; hence \(\log_{4}256=4\).
Set \(\log_{10}0.000001 = x\). Then \(\log_{10}10^{-6} = x\). Rewritten in exponential form, \(10^{x} = 10^{-6}\). Since the exponential function is a one-to-one function, \(x = -6\); hence \(\log_{10}0.000001 = -6\).
Set $ {27}=x $ . Rewritten in exponential form, $ 27^{x}= $ , or $ (3{3}){x}=3{3x}=3{-2} $ . Since the exponential function is a one-to-one function, $ 3x=-2 $ , $ x=- $ , hence $ {27}=- $ .
Set $ {1/5}125 = x $ . Rewritten in exponential form, $ ()^{x} = 125 $ , or $ (5{-1}){x} = 5^{-x} = 5^{3} $ . Since the exponential function is a one-to-one function, -x = 3, x = -3; hence $ {1/5}125 = -3 $ .
18.4. (a) Determine the domain and range of the logarithm function to base a.
Evaluate $ _{5}(-25) $ .
Since the logarithm function is the inverse function to the exponential function with base \(a\), and since the exponential function has domain \(R\) and range \((0,\infty)\), the logarithm function must have domain \((0,\infty)\) and range \(R\).
Since -25 is not in the domain of the logarithm function, $ _{5}(-25) $ is undefined.
18.5. Sketch a graph of $ f(x) = a^{x} $ , $ a > 1 $ , $ f^{-1}(x) = _{a} x $ , and the line y = x on the same Cartesian coordinate system
Note: The graph is shown in Fig. 18-1.
The domain of f is R and the range of f is (0,∞).
The points $ (0,1) $ and $ (1,a) $ are on the graph of f.
The negative x axis is an asymptote.
The domain of $ f^{-1} $ is $ (0,) $ and the range of $ f^{-1} $ is R.
The points $ (1,0) $ and $ (a,1) $ are on the graph of $ f^{-1} $ .
The negative y axis is an asymptote.
18.6. Sketch a graph of
$ f(x) = _{5} x $
Form a table of values.
$ g(x) = _{1/4} x $
| x | y |
| $ $ | -1 |
| 1 | 0 |
| 5 | 1 |
| 25 | 2 |
- Form a table of values.
| x | y |
| $ $ | 1 |
| 1 | 0 |
| 4 | -1 |
| 16 | -2 |
Domain: $ (0,) $ , range: R
Asymptote: negative y-axis.
The graph is shown in Fig. 18-2.
Domain: $ (0,) $ , range: R
Asymptote: positive y-axis.
The graph is shown in Fig. 18-3.
18.7. Prove the logarithmic–exponential function relations.
If $ y = {a} x $ , then $ x = a^{y} $ . Hence $ x = a^{y} = a^{{a} x} $ .
Similarly, reversing the letters, if $ x = {a} y $ , then $ y = a^{x} $ . Hence $ x = {a} y = _{a} a^{x} $ .
18.8. Show that if $ {a} u = {a} v $ , then u = v
Since the exponential function, $ f(x) = a^{x} $ , is a one-to-one function, its inverse function, $ f^{-1}(x) = {a} x $ , is also a one-to-one function and $ (f f^{-1})(x) = f(f^{-1}(x)) = f({a} x) = a^{_{a} x} = x $ .
Hence, if $ {a}u = {a}v $ , then $ a^{{a}u} = a^{{a}v} $ and u = v.
18.9. Evaluate, using the logarithmic–exponential function relations:
$ {3}3^{5} $ ; (b) $ {2}256 $ ; (c) $ _{a} $ ; (d) $ $ ;
$ 5^{{5}3} $ ; (f) $ e^{} $ ; (g) $ a{_{a}(x{2}-5x+6)} $ ; (h) $ 36^{{6}7} $
Ans. (a) $ {3}3^{5}=5 $ ; (b) $ {2}256=_{2}2^{8}=8 $ ;
$ {a}={a}a^{2/3}= $ ; (d) $ =_{10}10^{-5}=-5 $ ;
$ 5^{{5}3} = 3 $ ; (f) $ e^{} = e^{{e}} = $ ;
\[ \left(\mathrm{g}\right)a^{\log_{a}\left(x^{2}-5x+6\right)}=x^{2}-5x+6;\\\left(\mathrm{h}\right)36^{\log_{6}7}=\left(6^{2}\right)^{\log_{6}7}=6^{2\log_{6}7}=\left(6^{\log_{6}7}\right)^{2}=7^{2}=49 \]
18.10. Prove the properties of logarithms
The properties \(\log_{a}1=0\) and \(\log_{a}a=1\) follow directly from the logarithmic–exponential relations, since \(\log_{a}1=\log_{a}a^{0}=0\) and \(\log_{a}a=\log_{a}a^{1}=1\).
To prove the other properties, let $ u = {a} M $ and $ v = {a} N $ . Then $ M = a^{u} $ and $ N = a^{v} $ .
Therefore $ MN = a{u}a{v} $ ; thus $ MN = a^{u+v} $ . Rewriting in logarithmic form, $ _{a}MN = u + v $ .
Hence log a MN = log a M + log a N.
Similarly, $ = $ , thus $ = a^{u-v} $ . Rewriting in logarithmic form, $ _{a} = u - v $ .
Hence $ {a}()={a}M-_{a}N. $
Finally $ M{p}=(a{u}){p}=a{up} $ ; thus $ M{p}=a{pu} $ . Rewriting in logarithmic form, $ {a}M^{p}=pu $ . Hence $ {a}M^{p}=p_{a}M $ .
18.11. Use the properties of logarithms to rewrite in terms of logarithms of simpler expressions:
\[ \mathrm{o g}_{a}\frac{x y}{z};(\mathrm{b})\log_{a}(x^{2}-1);(\mathrm{c})\log_{a}\frac{x^{3}(x+5)}{(x-4)^{2}};(\mathrm{d})\log_{a}\sqrt{\frac{x^{2}+y^{2}}{x y}};(\mathrm{e})\ln(C e^{5x+1}) \]
Ans. (a) $ {a}={a}xy-{a}z={a}x+{a}y-{a}z $
\(\log_{a}(x^{2}-1)=\log_{a}[(x-1)(x+1)]=\log_{a}(x-1)+\log_{a}(x+1)\) Note: The properties of logarithms can be used to transform expressions involving logarithms of products, quotients, and powers. They do not allow simplification of logarithms of sums or differences.
$ {a}={a}x{3}+{a}(x+5)-{a}(x-4){2}=3{a}x+{a}(x+5)-2_{a}(x-4) $
\[ \left(\mathrm{d}\right)\log_{a}\sqrt{\frac{x^{2}+y^{2}}{xy}}=\frac{1}{2}\left[\log_{a}\frac{x^{2}+y^{2}}{xy}\right]=\frac{1}{2}\left[\log_{a}(x^{2}+y^{2})-\log_{a}(xy)\right]=\frac{1}{2}\left[\log_{a}(x^{2}+y^{2})-\log_{a}x-\log_{a}y\right] \]
\[ \mathrm{(e)}\ln(Ce^{5x+1})=\ln C+\ln e^{5x+1}=\ln C+5x+1 \]
18.12. Write as one logarithm:
- $ 3 {a} u - {a} v $ ; (b) $ {a} 5 - 3 {a} x - 4 {a} y $ ; (c) $ {a} (x - 3) + 3 {a} x + 2 {a} (1 + x) $ ;
\[ \begin{array}{l}\frac{1}{2}[\log_{a}x+3\log_{a}y-5\log_{a}(z-2)];\left(e\right)\frac{1}{2}\ln(x+1)-\frac{1}{2}\ln\left(x-1\right)+\ln C\end{array} \]
Ans. (a) $ 3 {a} u - {a} v = {a} u^{3} - {a} v = _{a} $
\[ \left(\mathrm{b}\right)\frac{1}{3}\log_{a}5-3\log_{a}x-4\log_{a}y=\log_{a}\bigvee^{3}\sqrt{5}-\log_{a}\left(x^{3}y^{4}\right)=\log_{a}\frac{\bigvee^{3}\sqrt{5}}{x^{3}y^{4}} \]
\[ \begin{aligned}(c)\ \frac{1}{3}\log_{a}{(x-3)}+3\log_{a}{x+2\log_{a}{(1+x)}}&=\log_{a}{(x-3)^{1/3}}+\log_{a}{x^{3}}+\log_{a}{(1+x)^{2}}\\&=\log_{a}{\sqrt[3]{x-3}}+\log_{a}{x^{3}(1+x)^{2}}\\&=\log_{a}{[x^{3}(1+x)^{2}\sqrt[3]{x-3}]}\end{aligned} \]
\[ \begin{aligned}(d)\frac{1}{2}\left[\log_{a}x+3\log_{a}y-5\log_{a}(z-2)\right]=\frac{1}{2}\left[\log_{a}xy^{3}-\log_{a}(z-2)^{5}\right]\end{aligned} \]
\[ \begin{aligned}(e)\ \frac{1}{2}\ln(x+1)-\frac{1}{2}\ln(x-1)+\ln C&=\frac{1}{2}\ln\left(\frac{x-1}{x+1}\right)+\ln C\\&=\ln\sqrt{\frac{x-1}{x+1}}+\ln C=\ln C\sqrt{\frac{x-1}{x+1}}\end{aligned} \]
SUPPLEMENTARY PROBLEMS
18.13. Write in exponential form:
\[ \log_{1000}10=\frac{1}{3};(b)\log_{7}\frac{1}{49}=-2;(c)\log_{u}\frac{1}{\sqrt{u}}=-\frac{1}{2} \]
Ans. (a) $ 1000^{1/3} = 10 $ ; (b) $ 7^{-2} = $ ; (c) $ u^{-1/2} = $
18.14. Write in logarithmic form:
- $ ()^{-3}=64 $ ; (b) $ e{2/5}=\sqrt[5]{e{2}} $ ; (c) $ m^{-p}=T $
Ans. (a) $ {1/4}64 = -3 $ ; (b) $ = $ ; (c) $ {m}T = -p $
18.15. Evaluate:
- $ Ce^{-at} $ ; (b) $ {4} $ ; (c) $ {10}(-100) $ ; (d) $ _{1/256} $
Ans. (a) $ C - at $ ; (b) $ - $ ; (c) undefined; (d) $ $
18.16. (a) Evaluate $ {3}81 $ ; (b) Evaluate $ {3} $ ; (c) Show that $ {a} = -{a}N $ . Ans. (a) 4; (b) -4
18.17. (a) Evaluate $ {5}125 $ ; (b) Evaluate $ {125}5 $ ; (c) Show that $ _{a}b= $
Ans. (a) 3; (b) $ $
18.18. Write in terms of logarithms of simpler expressions:
- $ {a}a(x-r)(x-s) $ ; (b) $ {a} $ ; (c) $ $ ; (d) $ $
Ans. (a) $ 1 + {a}(x - r) + {a}(x - s) $ ; (b) $ 2 - 3 {a}x - 4 {a}y $
- $ (a+)-(a-) $ , or (after rationalizing the denominator)
\[ 2 \ln (a + \sqrt{a^2 - x^2}) - 2 \ln x; \]
- $ (e{x}-e{-x})- $
18.19. Evaluate (a) $ 10^{(1/2)} $ ; (b) $ 5^{3{5}7} $ ; (c) $ 2^{-3{2}5} $
Ans. (a) $ $ ; (b) 343; (c) $ $
18.20. Write as one logarithm:
- $ 2 x - 8 y + 4 z $ ; (b) $ (1 - x) + (x - 3) $ ;
\[ )\frac{\ln\left(x+h\right)-\ln x}{h};\left(\mathrm{d}\right)x\ln x-\left(x-1\right)\ln(x-1); \]
Ans. (a) $ $ ; (b) undefined (there is no value of x for which both logarithms are defined);
\[ \left(\mathrm{c}\right)\ln\left(\frac{x+h}{x}\right)^{1/h};\left(\mathrm{d}\right)\ln\frac{x^{x}}{(x-1)^{x-1}};\left(\mathrm{e}\right)1-\log_{c}a \]
18.21. Given log a = 0.69, log b = 1.10, and log a = 1.61, use the properties of logarithms to evaluate:
\[ \left(\mathrm{a}\right)\log_{a}30;\left(\mathrm{b}\right)\log_{a}\frac{6}{5};\left(\mathrm{c}\right)\log_{a}\frac{1}{\sqrt{15}};\left(\mathrm{d}\right)\log_{a}\left(-\frac{5}{6}\right) \]
Ans. (a) 3.40; (b) 0.18; (c) -1.36; (d) undefined
18.22. Sketch graphs of:
\[ \begin{array}{r l}&{\mathrm{(a)~}f(x)=\log_{3}(x+2);\mathrm{(b)~}F(x)=3-\log_{2}x;\mathrm{(c)~}g(x)=\ln\left\vert x\right\vert;\mathrm{(d)~}G(x)=-\ln{(-x)};}\end{array} \]
Ans. (a) Fig. 18-4; (b) Fig. 18-5; (c) Fig. 18-6; (d) Fig. 18-7
