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Mathematics · Ch 15 — Functions

Logarithmic Function

15.1.5.7

Logarithmic Function

8. Logarithmic function. Let a>0, a≠1a>0,\ a\ne1; we define y=log⁡axy=\log_ax if x=ayx=a^y, for x>0x>0. This gives two equivalent forms: the logarithmic form y=log⁡axy=\log_ax and the exponential form ay=xa^y=x — each converts directly to the other.

Properties: (1) Since a0=1a^0=1, log⁡a1=0\log_a1=0; and since a1=aa^1=a, log⁡aa=1\log_aa=1. (2) Since ax=ay  ⟺  x=ya^x=a^y\iff x=y, so log⁡ax=log⁡ay  ⟺  x=y\log_ax=\log_ay\iff x=y. (3) Product rule: for a,b,c>0a,b,c>0 and a≠1a\ne1, log⁡a(bc)=log⁡ab+log⁡ac\log_a(bc)=\log_ab+\log_ac (verify!). (4) Quotient rule: for a,b,c>0a,b,c>0 and a≠1a\ne1, log⁡abc=log⁡ab−log⁡ac\log_a\dfrac{b}{c}=\log_ab-\log_ac (verify!). (5) Power/exponent rule: for a,b,c>0a,b,c>0 and a≠1a\ne1, log⁡abc=clog⁡ab\log_ab^c=c\log_ab (verify!). (6) For the natural base ee, log⁡ex=ln⁡x\log_ex=\ln x, the natural logarithm function (Fig. 6.30). Domain of ln⁡x\ln x: (0,∞)(0,\infty); Range: (−∞,∞)(-\infty,\infty).

Logarithmic inequalities: (i) If a>1a>1 and 0<m<n0<m<n, then log⁡am<log⁡an\log_am<\log_an (e.g. log⁡1020<log⁡1030\log_{10}20<\log_{10}30). (ii) If 0<a<10<a<1 and 0<m<n0<m<n, then log⁡am>log⁡an\log_am>\log_an (e.g. log⁡0.120>log⁡0.130\log_{0.1}20>\log_{0.1}30). (iii) For a,m>0a,m>0, if aa and mm lie on the same side of 11, then log⁡am>0\log_am>0 (e.g. log⁡23>0\log_23>0, log⁡0.30.5>0\log_{0.3}0.5>0). (iv) For a,m>0a,m>0, if aa and mm lie on different sides of 11, then log⁡am<0\log_am<0 (e.g. log⁡0.23<0\log_{0.2}3<0, log⁡30.5<0\log_30.5<0).

Ex. 11: Write log⁡72\log72 in terms of log⁡2\log2 and log⁡3\log3.

Solution: log⁡72=log⁡(23⋅32)=log⁡23+log⁡32\log72=\log(2^3\cdot3^2)=\log2^3+\log3^2 (product rule) =3log⁡2+2log⁡3=3\log2+2\log3 (power rule).

Ex. 12: Evaluate ln⁡e9−ln⁡e4\ln e^9-\ln e^4.

Solution: ln⁡e9−ln⁡e4=log⁡ee9−log⁡ee4=9log⁡ee−4log⁡ee=9(1)−4(1)=5\ln e^9-\ln e^4=\log_ee^9-\log_ee^4=9\log_ee-4\log_ee=9(1)-4(1)=5 (since ln⁡e=1\ln e=1).

Ex. 13: Expand log⁡x3(x+3)2(x−4)2\log\dfrac{x^3(x+3)}{2(x-4)^2}.

Solution: Using the quotient rule: =log⁡[x3(x+3)]−log⁡[2(x−4)2]=\log[x^3(x+3)]-\log[2(x-4)^2]. Using the product rule: =[log⁡x3+log⁡(x+3)]−[log⁡2+log⁡(x−4)2]=[\log x^3+\log(x+3)]-[\log2+\log(x-4)^2]. Using the power rule: =3log⁡x+log⁡(x+3)−log⁡2−2log⁡(x−4)=3\log x+\log(x+3)-\log2-2\log(x-4).

Ex. 14: Combine 3ln⁡(p+1)−12ln⁡r+5ln⁡(2q+3)3\ln(p+1)-\dfrac12\ln r+5\ln(2q+3) into a single logarithm.

Solution: Using the power rule: =ln⁡(p+1)3−ln⁡r1/2+ln⁡(2q+3)5=\ln(p+1)^3-\ln r^{1/2}+\ln(2q+3)^5. Using the quotient rule (on the first two terms): =ln⁡(p+1)3r1/2+ln⁡(2q+3)5=\ln\dfrac{(p+1)^3}{r^{1/2}}+\ln(2q+3)^5. Using the product rule: =ln⁡[(p+1)3(2q+3)5r1/2]=\ln\left[\dfrac{(p+1)^3(2q+3)^5}{r^{1/2}}\right].

Ex. 15: Find the domain of ln⁡(x−5)\ln(x-5).

Solution: ln⁡(x−5)\ln(x-5) is defined for x−5>0x-5>0, that is x>5x>5, so the domain is (5,∞)(5,\infty). …

Figure 1Fig. 6.30 — graph of the natural logarithm f(x)=ln(x)

What this figure shows. A curve starting from far down near the y-axis (as x approaches 0 from the right, the curve plunges toward negative infinity), crossing the x-axis at (1,0), and rising slowly and steadily to the right, passing through roughly (e,1) and continuing to climb without bound but ever more slowly. Illustrates domain (0,infinity) and range R (all reals), the mirror-image relationship to the exponential curve, and that the logarithm grows unboundedl …

Figure 2Fig. 6.31 — a number-line domain diagram for Ex. 18's f(x)=log base (x+5) of (x^2-4)

What this figure shows. A horizontal number line marked with the integers from -6 to 4, with several overlapping shaded bands and open/closed circles drawn above it representing each separate condition needed for the function to be defined: a shaded region for x<-2 or x>2 (from the argument x^2-4>0), a shaded ray for x>-5 (from the base x+5>0), and a small excluded open circle at x=-4 (from the base x+5 not equal to 1). The diagram is used to visually intersect all three conditions and read off the combined domain …