# Section 4.5 Exponential and Logarithmic Equations

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1 Section 4.5 Exponential and Logarithmic Equations Exponential Equations An exponential equation is one in which the variable occurs in the exponent. EXAMPLE: Solve the equation x = 7. Solution 1: We have x = 7 log x = log 7 ] xlog = log 7 ] Solution : We have EXAMPLE: Solve the equation 4 x+1 = 3. Solution 1: We have x = log x = 7 ln x = ln7 xln = ln7 x = ln7 ln x+1 = 3 log4 4 x+1 = log 4 3 ] (x+1)log4 4 = log 4 3 ] Solution : We have x+1 = log 4 3 x = log x+1 = 3 ln4 x+1 = ln3 (x+1)ln4 = ln3 EXAMPLE: Solve the equation 3 x 3 = 5. x+1 = ln3 ln4 x = ln ln4 1

2 EXAMPLE: Solve the equation 3 x 3 = 5. Solution 1: We have 3 x 3 = 5 x 3 = log 3 5 x = log Solution : We have 3 x 3 = 5 ln3 x 3 = ln5 (x 3)ln3 = ln5 x 3 = ln5 ln3 x = ln ln3 EXAMPLE: Solve the equation 8e x = 0. 8e x = 0 e x = 0 8 = 5 x = ln 5 x = ln 5 = 1 ln EXAMPLE: Solve the equation e 3 x = 4 algebraically and graphically. e 3 x = 4 3 x = ln4 x = ln4 3 x = ln4 3 = 3 ln

3 EXAMPLE: Solve the equation e x e x 6 = 0. Solution 1: We have e x e x 6 = 0 (e x ) e x 6 = 0 (e x 3)(e x +) = 0 e x 3 = 0 or e x + = 0 e x = 3 e x = The equation e x = 3 leads to x = ln3. But the equation e x = has no solution because e x > 0 for all x. Thus, x = ln is the only solution. Solution 1 : Put e x = w. Then e x e x 6 = 0 (e x ) e x 6 = 0 w w 6 = 0 (w 3)(w +) = 0 w 3 = 0 or w+ = 0 w = 3 e x = 3 w = e x = The equation e x = 3 leads to x = ln3. But the equation e x = has no solution because e x > 0 for all x. Thus, x = ln is the only solution. EXAMPLE: Solve the equation e x 3e x + = 0. 3

4 EXAMPLE: Solve the equation e x 3e x + = 0. Solution 1: We have e x 3e x + = 0 (e x ) 3e x + = 0 (e x 1)(e x ) = 0 e x 1 = 0 or e x = 0 e x = 1 e x = x = 0 x = ln Solution 1 : Put e x = w. Then w 3w + = 0 (w 1)(w ) = 0 w 1 = 0 or w = 0 w = 1 w = e x = 1 e x = x = 0 x = ln EXAMPLE: Solve the equation 7 x 3 7 x +1 = 0. Solution: Put 7 x = w. Then w 3w +1 = 0 = w = ( 3)± ( 3) ( so 7 x = 3± 5 3± ) 5, therefore x = log 7. = 3± 5 EXAMPLE: Solve the equation 7 x 7 x 1 = 0. Solution: Put 7 x = w. Then Since 1 5 w w 1 = 0 = w = ( 1)± ( 1) 4 1 ( 1) 1 < 0, it follows that 7 x = 1+ 5 = x = log 7 ( 1+ ) 5 = 1± 5 EXAMPLE: Solve the equation 3xe x +x e x = 0. 3xe x +x e x = 0 xe x (3+x) = 0 x(3+x) = 0 x = 0 or 3+x = 0 x = 0 x = 3 4

5 Logarithmic Equations A logarithmic equation is one in which a logarithm of the variable occurs. EXAMPLE: Solve the equation lnx = 8. lnx = 8 e lnx = e 8] x = e 8 EXAMPLE: Solve the equation log (x+) = 5. log (x+) = 5 log (x+) = 5] x+ = 5 x = 5 = 3 = 30 EXAMPLE: Solve the equation log 7 (5 x) = 3. log 7 (5 x) = 3 7 log (5 x) 7 = 7 3] 5 x = 7 3 x = = = 318 EXAMPLE: Solve the equation 4+3log(x) = 16. 5

6 EXAMPLE: Solve the equation 4+3log(x) = log(x) = 16 3log(x) = 1 log(x) = 4 x = 10 4 x = 104 = 10,000 = 5,000 EXAMPLE: Solve the equation log(x+)+log(x 1) = 1 algebraically and graphically. log(x+)+log(x 1) = 1 log(x+)(x 1)] = 1 (x+)(x 1) = 10 x +x = 10 x +x 1 = 0 (x+4)(x 3) = 0 x+4 = 0 or x 3 = 0 x = 4 x = 3 We check these potential solutions in the original equation and find that x = 4 is not a solution (because logarithms of negative numbers are undefined), but x = 3 is a solution. To solve the equation graphically we rewrite it as log(x+)+log(x 1) 1 = 0 and then graph y = log(x+)+log(x 1) 1. The solutions are the x-intercepts of the graph. EXAMPLE: Solve the following equations (a) log(x+8)+log(x 1) = 1 (b) log(x 1) log(x+1) = 3 6

7 EXAMPLE: Solve the following equations (a) log(x+8)+log(x 1) = 1 (b) log(x 1) log(x+1) = 3 Solution: (a) We have log(x+8)+log(x 1) = 1 log(x+8)(x 1)] = 1 (x+8)(x 1) = 10 x +7x 8 = 10 x +7x 18 = 0 (x+9)(x ) = 0 x+9 = 0 or x = 0 x = 9 x = We check these potential solutions in the original equation and find that x = 9 is not a solution (because logarithms of negative numbers are undefined), but x = is a solution. (b) We have log(x 1) log(x+1) = 3 log x 1 x+1 = 3 x 1 x+1 = 103 (x 1)(x+1) x+1 = 1000 x 1 = 1000 x = 1001 EXAMPLE: Solve the equation x = ln(x+) graphically. Solution: We first move all terms to one side of the equation x ln(x+) = 0. Then we graph y = x ln(x+). The solutions are the x-intercepts of the graph. EXAMPLE: Find the solution of the equation, correct to two decimal places. (a) 10 x+3 = 6 x (b) 5 ln(3 x) = 4 (c) log (x+)+log (x 1) = 7

8 EXAMPLE: Find the solution of the equation, correct to two decimal places. (a) 10 x+3 = 6 x (b) 5 ln(3 x) = 4 (c) log (x+)+log (x 1) = Solution: (a) We have 10 x+3 = 6 x ln10 x+3 = ln6 x (x+3)ln10 = xln6 xln10+3ln10 = xln6 xln10 xln6 = 3ln10 x(ln10 ln6) = 3ln10 x = 3ln10 ln10 ln (b) We have 5ln(3 x) = 4 ln(3 x) = x = e 4/5 x = 3 e 4/ (c) We have log (x+)+log (x 1) = log (x+)(x 1) = (x+)(x 1) = 4 x +x = 4 x +x 6 = 0 (x )(x+3) = 0 x = 0 or x+3 = 0 x = x = 3 Since x = 3 is not from the domain of log (x+)+log (x 1), the only answer is x =. 8

9 Applications EXAMPLE: If I 0 and I denote the intensity of light before and after going through a material and x is the distance (in feet) the light travels in the material, then according to the Beer- Lambert Law 1 k ln ( I I 0 ) = x where k is a constant depending on the type of material. (a) Solve the equation for I. (b) For a certain lake k = 0.05 and the light intensity is I 0 = 14 lumens (lm). Find the light intensity at a depth of 0 ft. Solution: (a) We first isolate the logarithmic term. 1 k ln ( I I 0 ) = x ( ) I ln = kx I 0 I I 0 = e kx I = I 0 e kx (b) We find I using the formula from part (a). I = I 0 e kx = 14e ( 0.05)(0) 8.49 The light intensity at a depth of 0 ft is about 8.5 lm. EXAMPLE: A sum of \$5000 is invested at an interest rate of 5% per year. Find the time required for the money to double if the interest is compounded according to the following method. (a) Semiannual (b) Continuous 9

10 EXAMPLE: A sum of \$5000 is invested at an interest rate of 5% per year. Find the time required for the money to double if the interest is compounded according to the following method. (a) Semiannual (b) Continuous Solution: (a) We use the formula for compound interest A(t) = P ( 1+ r ) nt n with P = \$5000, A(t) = \$10,000, r = 0.05, n = and solve the resulting exponential equation for t. ( ) t = 10,000 The money will double in years. (1.05) t = log1.05 t = log tlog1.05 = log t = log log (b) We use the formula for continuously compounded interest A(t) = Pe rt with P = \$5000, A(t) = \$10,000, r = 0.05 and solve the resulting exponential equation for t. 5000e 0.05t = 10,000 e 0.05t = 0.05t = ln The money will double in years. t = ln EXAMPLE: A sum of \$1000 is invested at an interest rate of 4% per year. Find the time required for the amount to grow to \$4000 if interest is compounded continuously. 10

11 EXAMPLE: A sum of \$1000 is invested at an interest rate of 4% per year. Find the time required for the amount to grow to \$4000 if interest is compounded continuously. Solution: We use the formula for continuously compounded interest A(t) = Pe rt with P = \$1000, A(t) = \$4000, r = 0.04 and solve the resulting exponential equation for t. 1000e 0.04t = 4000 e 0.04t = t = ln4 t = ln The amount will be \$4000 in about 34 years and 8 months. 11

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