creatures contain the same fixed proportions of 14( to nonradioactive 12C as the atmosphere. After an organism dies, it stops assimilating 14(, and the amount of 14( in it begins to decay exponentially. We can then determine the time elapsed since the death of the organism by measuring the amount of 14( left in it. For example, if a donkey bone contains 73% as much 14( as a living donkey and it died t years ago, then by the formula for radioactive de-cay (Section 4.6), 0.73 = (l.OO)e-(tln2)/5730 We solve this exponential equation to find t = 2600, so the bone is about 2600 years old. • •

334 CHAPTER 4 I Exponential and Logarithmic Functions CHECK YOUR ANSWER lfx = 17, we get log2(25-17) = log28 = 3 V l' Logarithmic Equations A lugarirlunic equation is one in which a logarithm of the variable occurs. For example, logl r + 2) = 5 To solve for x. we write the equation in exponential form. E\ponential rorm X= 32-2 = 30 Solve for x Another way of looking at the first step is to raise the base. 2, to each side of the equation. Rui-.e 2 to each -,ide Propert) or logarithms X= 32-2 = 30 Solve for .I The method used to solve this simple problem is typical. We summarize the steps as follows. GUIDELINES FOR SOLVING LOGARITHMIC EQUATIONS 1. Isolate the logarithmic term on one side of the equation; you might lirst need to combine the logarithmic terms. 2. Write the equation in exponential form (or raise the base to each side of the equation). 3. Solve for the variable. EXAMPLE 6 I Solving Logarithmic Equations Solve each equation for x. (a) In x = 8 (b) log2(25 -x) = 3 SOLUTION (a) In .r = 8 Therefore, .r = e8 = 298 1. Gi,·en equation Exponential form We can also solve this problem another way: In .r = 8 Gi1cn equation Rai~e e to each -,ide Propcrt) of In (b) The first step is to rewrite the equation in exponential form. Gi1cn equation 25 -.r = 2' c\poncntial form (or raise 2 to each -.ide) 25 -.r = 8 X= 25-8 = 17 •, NOW TRY EXERCISES 37 AND 41 •

CHECK YOUR ANSWER x = - 4: log( - 4 + 2) + log( - 4 -I ) = log( -2) + log( - 5) undefined X .r = 3: log(3 + 2) + log(3 -I) 3 = log 5 + log 2 = log(5 · 2) = log 10 = I V' 0 ----+r----+-=--1--==;=::--:;:=:==t----+---6 -3 FIGURE 2 SECTION 4.5 I Exponential and Logarithmic Equations 335 EXAMPLE 7 I Solving a Logarithmic Equation Solve the equation 4 + 3 log(2x) = 16. S 0 LUTI 0 N We flrst isolate the logarithmic term. This allows us to write the equation in exponential form. 4 + 3 log(2x) = 16 3 log(2x) = 12 log(2x) = 4 2x = 10~ X= 5000 CHECK YOUR ANSWER If x = 5000, we get Gt1·en equation Subtract -1 Divid..: by J E\pon..:ntial form (or raise I0 10 ea<.:h ' ide) Di1 ide hy:! 4 + 3 log 2(5000) = 4 + 3 log 10,000 = 4 + 3(4) = 16 V' •. NOW TRY EXERCISE 43 EXAMPLE 8 I Solving a Logarithmic Equation Algebraically and Graphically Solve the equation log(.r + 2) + log(x-I ) = I algebraically and graphically. SOLUTION 1: Algebraic We flrst combine the logarithmic terms, using the Laws of Logarithms.