Solving Exponential and Logarithmic Equations Section 4.4 JMerrill, 2005 Revised, 2008
Same Base SSSSolve: 4x-2 = 64x 4 x-2 = (43)x x-2 = 43x x –2 = 3x - 2 = 2x 1 = x If b M = b N, then M = N 64 = 4 3 If the bases are already =, just solve the exponents
You Do Solve 27 x+3 = 9 x-1
Review – Change Logs to Exponents llllog3x = 2 llllogx16 = 2 llllog 1000 = x 3 2 = x, x = 9 x 2 = 16, x = 4 10 x = 1000, x = 3
Using Properties to Solve Logarithmic Equations IIIIf the exponent is a variable, then take the natural log of both sides of the equation and use the appropriate property. TTTThen solve for the variable.
Example: Solving 2 x = 7 problem lllln2x = ln7 take ln both sides xxxxln2 = ln7power rule x = divide to solve for x = 2.807
Example: Solving e x = 72problem l nex = ln 72take ln both sides x lne = ln 72power rule = 4.277solution: because ln e = ?
You Do: Solving 2 ex + 8 = 20problem ex = 12subtract 8 e x = 6divide by 2 l n ex = ln 6take ln both sides x lne = 1.792power rule x = 1.792(remember: lne = 1)
Example SSSSolve 5x-2 = 42x+3 lllln5x-2 = ln42x+3 ((((x-2)ln5 = (2x+3)ln4 TTTThe book wants you to distribute… IIIInstead, divide by ln4 ((((x-2) = 2x+3 1111.1609x = 2x+3 xxxx≈6.3424
Solving by Rewriting as an Exponential SSSSolve log4(x+3) = 2 44442 = x+3 11116 = x+3 11113 = x
You Do SSSSolve 3ln(2x) = 12 lllln(2x) = 4 RRRRealize that our base is e, so eeee4 = 2x xxxx ≈ YYYYou always need to check your answers because sometimes they don’t work!
Using Properties to Solve Logarithmic Equations 1111.Condense both sides first (if necessary). 2222.If the bases are the same on both sides, you can cancel the logs on both sides. 3333.Solve the simple equation
Example: Solve for x llllog36 = log33 + log3xproblem llllog36 = log33xcondense 6 = 3xdrop logs 2 = xsolution
You Do: Solve for x llllog 16 = x log 2problem llllog 16 = log 2xcondense 1 6 = 2xdrop logs x = 4solution
You Do: Solve for x l og4x = log44problem = log44condense 4drop logs ccccube each side X = 64solution
Example 7xlog 2 5 = 3xlog ½ log 2 25 log 2 5 7x = log 2 5 3x + log 2 25 ½ log 2 5 7x = log 2 5 3x + log 7x = 3x + 1 4x = 1
You Do Solve:log log 7 2 = log 7 x + log 7 (5x – 3)
You Do Answer Solve:log log 7 2 = log 7 x + log 7 (5x – 3) log 7 14 = log 7 x(5x – 3) 14 = 5x 2 -3x 0 = 5x 2 – 3x – 14 0 = (5x + 7)(x – 2) Do both answers work? NO!!
Final Example How long will it take for $25,000 to grow to $500,000 at 9% annual interest compounded monthly?
Example