Solving equations involving exponents and logarithms

Let’s review some terms Let’s review some terms. When we write log5 125 5 is called the base 125 is called the argument

Logarithmic form of 52 = 25 is log525 = 2

For all the laws a, M and N > 0 r is any real

ln is a short cut for loge log means log10 Remember ln and log ln is a short cut for loge log means log10

Log laws

If your variable is in an exponent or in the argument of a logarithm Find the pattern your equation resembles

If your variable is in an exponent or in the argument of a logarithm Find the pattern Fit your equation to match the pattern Switch to the equivalent form Solve the result Check (be sure you remember the domain of a log)

log(2x) = 3

log(2x) = 3 It fits

log(2x) = 3 103=2x Switch Did you remember that log(2x) means log10(2x)?

log(2x) = 3 103=2x 500 = x Divide by 2

ln(x+3) = ln(-7x)

ln(x+3) = ln(-7x) It fits

ln(x+3) = ln(-7x) Switch

ln(x+3) = ln(-7x) x + 3 = -7x Switch

ln(x+3) = ln(-7x) x + 3 = -7x x = - ⅜ Solve the result (and check)

ln(x) + ln(3) = ln(12)

ln(x) + ln(3) = ln(12) x + 3 = 12

ln(x) + ln(3) = ln(12) x + 3 = 12 Oh NO!!! That’s wrong!

ln(x) + ln(3) = ln(12) ln(3x) = ln (12) You need to use log laws

ln(x) + ln(3) = ln(12) ln(3x) = ln (12) 3x = 12 Switch

ln(x) + ln(3) = ln(12) ln(3x) = ln (12) 3x = 12 x = 4 Solve the result

log3(x+2) + 4 = 9

log3(x+2) + 4 = 9 It will fit

log3(x+2) + 4 = 9 log3(x+2) = 5 Subtract 4 to make it fit

log3(x+2) + 4 = 9 log3(x+2) = 5 Switch

log3(x+2) + 4 = 9 log3(x+2) = 5 35 = x + 2 Switch

log3(x+2) + 4 = 9 log3(x+2) = 5 35 = x + 2 x = 241 Solve the result

5(10x) = 19.45

5(10x) = 19.45 10x = 3.91 Divide by 5 to fit

5(10x) = 19.45 10x = 3.91 Switch

5(10x) = 19.45 10x = 3.91 log(3.91) = x Switch

5(10x) = 19.45 10x = 3.91 log(3.91) = x ≈ 0.592 Exact log(3.91) Approx 0.592

2 log3(x) = 8

2 log3(x) = 8 It will fit

2 log3(x) = 8 log3(x) = 4 Divide by 2 to fit

2 log3(x) = 8 log3(x) = 4 Switch

2 log3(x) = 8 log3(x) = 4 34=x Switch

2 log3(x) = 8 log3(x) = 4 34=x x = 81 Then Simplify

log2(x-1) + log2(x-1) = 3

log2(x-1) + log2(x-1) = 3 Need to use a log law

log2(x-1) + log2(x+1) = 3 log2{(x-1)(x+1)} = 3

log2(x-1) + log2(x+1) = 3 log2{(x-1)(x+1)} = 3 Switch

log2(x-1) + log2(x+1) = 3 log2{(x-1)(x+1)} = 3 23=(x-1)(x+1) Switch

log2(x-1) + log2(x+1) = 3 log2{(x-1)(x+1)} = 3 23=(x-1)(x+1) = x2 -1 x = +3 or -3 and finish

log2(x-1) + log2(x+1) = 3 log2{(x-1)(x+1)} = 3 23=(x-1)(x+1) = x2 -1 x = +3 or -3 But -3 does not check!

Exclude -3 (it would cause you to have a negative argument) log2(x-1) + log2(x+1) = 3 log2{(x-1)(x+1)} = 3 23=(x-1)(x+1) = x2 -1 x = +3 or -3 Exclude -3 (it would cause you to have a negative argument)

There’s more than one way to do this

Can you find why each step is valid?

rules of exponents multiply both sides by 2 - 3 to get exact answer Approximate answer

Here’s another way to solve the same equation.

Square both sides Simplify exclude 2nd result

52x - 5x – 12 = 0

Factor it. Think of y2 - y-12=0 52x - 5x – 12 = 0 (5x – 4)(5x + 3) = 0 Factor it. Think of y2 - y-12=0

52x - 5x – 12 = 0 (5x – 4)(5x + 3) = 0 5x – 4 = 0 or 5x + 3 = 0 Set each factor = 0

Solve first factor’s equation Solve 5x – 4 = 0 5x = 4 log54 = x Solve first factor’s equation

Solve other factor’s equation Solve 5x + 3 = 0 5x = -3 log5(-3) = x Solve other factor’s equation

Oops, we cannot have a negative argument Solve 5x + 3 = 0 5x = -3 log5(-3) = x Oops, we cannot have a negative argument

Only the other factor’s solution works Solve 5x + 3 = 0 5x = -3 log5(-3) = x Exclude this solution. Only the other factor’s solution works

4x+2 = 5x

4x+2 = 5x If M = N then ln M = ln N

4x+2 = 5x ln(4x+2) = ln(5x ) If M = N then ln M = ln N

4x+2 = 5x ln(4x+2) = ln(5x)

4x+2 = 5x ln(4x+2) = ln(5x) (x+2)ln(4) =(x) ln(5 )

Distribute 4x+2 = 5x ln(4x+2) = ln(5x) (x+2)ln(4) =(x) ln(5 ) x ln (4) + 2 ln(4) = x ln (5) Distribute

Get x terms on one side 4x+2 = 5x ln(4x+2) = ln(5x) x ln (4) + 2 ln(4) = x ln (5) x ln (4) - x ln (5)= 2 ln(4) Get x terms on one side

Factor out x 4x+2 = 5x ln(4x+2) = ln(5x) (x+2)ln(4) =(x) ln(5 ) x ln (4) + 2 ln(4) = x ln (5) x ln (4) - x ln (5)= 2 ln(4) x [ln (4) - ln (5)]= 2 ln(4) Factor out x

Divide by numerical coefficient 4x+2 = 5x ln(4x+2) = ln(5x) (x+2)ln(4) =(x) ln(5 ) x ln (4) + 2 ln(4) = x ln (5) x ln (4) - x ln (5)= 2 ln(4) x [ln (4) - ln (5)]= 2 ln(4) Divide by numerical coefficient