7.6 Solve Exponential and Logarithmic Equations

Exponential Equations One way to solve exponential equations is to use the property that if 2 powers w/ the same base are equal, then their exponents are equal. For b>0 & b≠1 if bx = by, then x=y

Solve by equating exponents 43x = 8x+1 (22)3x = (23)x+1 rewrite w/ same base 26x = 23x+3 6x = 3x+3 x = 1 Check → 43*1 = 81+1 64 = 64

24x = 32x-1 24x = (25)x-1 4x = 5x-5 5 = x Your turn: Be sure to check your answer!!!

When you can’t rewrite using the same base, you can solve by taking a log of both sides 2x = 7 log22x = log27 x = log27 x = ≈ 2.807

4x = 15 log44x = log415 x = log415 = log15/log4 ≈ 1.95

102x-3+4 = 21 -4 -4 102x-3 = 17 log10102x-3 = log1017 2x-3 = log 17 -4 -4 102x-3 = 17 log10102x-3 = log1017 2x-3 = log 17 2x = 3 + log17 x = ½(3 + log17) ≈ 2.115

5x+2 + 3 = 25 5x+2 = 22 log55x+2 = log522 x+2 = log522 = (log22/log5) – 2 ≈ -.079

If logbx = logby, then x = y Solving Log Equations To solve use the property for logs w/ the same base: + #’s b,x,y & b≠1 If logbx = logby, then x = y

5x – 1 = x + 7 5x = x + 8 4x = 8 x = 2 and check log3(5x-1) = log3(x+7) 5x – 1 = x + 7 5x = x + 8 4x = 8 x = 2 and check log3(5*2-1) = log3(2+7) log39 = log39

When you can’t rewrite both sides as logs w/ the same base exponentiate each side b>0 & b≠1 if x = y, then bx = by

3x+1 = 25 x = 8 and check log5(3x + 1) = 2 5log5(3x+1) = 52 Because the domain of log functions doesn’t include all reals, you should check for extraneous solutions

log5x + log(x+1)=2 10log5x -5x = 102 log (5x)(x+1) = 2 (product property) log (5x2 – 5x) = 2 10log5x -5x = 102 5x2 - 5x = 100 x2 – x - 20 = 0 (subtract 100 and divide by 5) (x-5)(x+4) = 0 x=5, x=-4 graph and you’ll see 5=x is the only solution 2

One More: log2x + log2(x-7) = 3 log2x(x-7) = 3 log2 (x2- 7x) = 3 2log2x -7x = 23 x2 – 7x = 8 x2 – 7x – 8 = 0 (x-8)(x+1)=0 x=8 x= -1 2