3 4 Solving Exponential and Logarithmic Equations Exponential

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3. 4 Solving Exponential and Logarithmic Equations

3. 4 Solving Exponential and Logarithmic Equations

Exponential Equations • One way to solve exponential equations is to use the property

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 x b = y b, then x=y

Solve by equating exponents 3 x • 4 x+1 8 = 2 3 x

Solve by equating exponents 3 x • 4 x+1 8 = 2 3 x 3 x+1 • (2 ) = (2 ) rewrite w/ same base 6 x 3 x+3 • 2 = 2 Check → 4 = 8 • 6 x = 3 x+3 64 = 64 • x = 1 3*1 1+1

Your turn! 4 x • 2 x-1 32 = 4 x 5 x-1 •

Your turn! 4 x • 2 x-1 32 = 4 x 5 x-1 • 2 = (2 ) • 4 x = 5 x-5 • 5 = x Be sure to check your answer!!!

When you can’t rewrite using the same base, you can solve by taking a

When you can’t rewrite using the same base, you can solve by taking a log or ln of both sides • 2 x = 7 x • log 22 = log 27 • x= ≈ 2. 807

x 4 = 15 Using Log • log 44 x = log 415 •

x 4 = 15 Using Log • log 44 x = log 415 • x = log 415 = log 15/log 4 • ≈ 1. 95 Using LN

2 x-3 10 +4 • • = 21 -4 -4 2 x-3 10 =

2 x-3 10 +4 • • = 21 -4 -4 2 x-3 10 = 17 2 x-3 log 1010 = log 1017 2 x-3 = log 17 2 x = 3 + log 17 x = ½(3 + log 17) ≈ 2. 115

Solving an Exponential Equation containing e • 40 e 0. 6 x-3= 237

Solving an Exponential Equation containing e • 40 e 0. 6 x-3= 237

Solving Log Equations • To solve use the property for logs w/ the same

Solving Log Equations • To solve use the property for logs w/ the same base use the one-to one property: • If logbx = logby, then x = y

log 3(5 x-1) = log 3(x+7) • 5 x – 1 = x +

log 3(5 x-1) = log 3(x+7) • 5 x – 1 = x + 7 • 5 x = x + 8 • 4 x = 8 • x = 2 and check • log 3(5*2 -1) = log 3(2+7) • log 39 = log 39

log 5(3 x + 1) = 2 Write in exponential form then solve! •

log 5(3 x + 1) = 2 Write in exponential form then solve! • 3 x+1 = 25 x = 8 and check • Because the domain of log functions doesn’t include all reals, you should check for extraneous solutions

log 5 x + log(x+1)=2 • log (5 x)(x+1) = 2 • log (5

log 5 x + log(x+1)=2 • log (5 x)(x+1) = 2 • log (5 x 2 – 5 x) = 2 • 5 x 2 - 5 x = 100 • (product property) x 2 – 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