Solving equations involving exponents and logarithms

Let’s review some terms.

When we write log

5 125

5 is called the base125 is called the argument

Logarithmic form of 52 = 25 is

log525 = 2

For all the lawsa, M and N > 0

a ≠ 1

r is any real

Remember ln and log

ln is a short cut for loge

log means log10

Log laws

a

MM

NMN

M

NMMN

MrM

a

a

aaa

aaa

ar

a

a

a

ln

lnlog

logloglog

logloglog

loglog

1log

01log

If your variable is in an exponent or in the argument of a logarithm

Find the pattern your equation resembles

NMNM

ennb be

lnln

log

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)

NMNM

ennb be

lnln

log

2

)5ln(x

log(2x) = 3

It fits

2

)5ln(x

log(2x) = 3

enb log

Switch

2

)5ln(x

log(2x) = 3

103=2x ennb be log

Did you remember that log(2x) means log10(2x)?

Divide by 2

2

)5ln(x

log(2x) = 3

103=2x

500 = x

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

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

It fitsNM lnln

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

Switch

NMNM lnln

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

x + 3 = -7x

Switch

NMNM lnln

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!

You need to use log laws

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

ln(3x) = ln (12)

Switch

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

ln(3x) = ln (12)

3x = 12

NMNM lnln

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

ln(3x) = ln (12)

3x = 12

x = 4 Solve the result

log3(x+2) + 4 = 9

It will fit

log3(x+2) + 4 = 9

enb log

Subtract 4 to make it fit

log3(x+2) + 4 = 9

log3(x+2) = 5

enb log

Switch

log3(x+2) + 4 = 9

log3(x+2) = 5

nben eb log

Switch

log3(x+2) + 4 = 9

log3(x+2) = 5

35 = x + 2

nben eb log

Solve the result

log3(x+2) + 4 = 9

log3(x+2) = 5

35 = x + 2

x = 241

5(10x) = 19.45

Divide by 5 to fit

5(10x) = 19.45

10x = 3.91

nbe

Switch

5(10x) = 19.45

10x = 3.91

ennb be log

Switch

5(10x) = 19.45

10x = 3.91

log(3.91) = x

ennb be log

Exact log(3.91)

Approx 0.592

5(10x) = 19.45

10x = 3.91

log(3.91) = x

≈ 0.592

2 log3(x) = 8

It will fit

2 log3(x) = 8

enb log

Divide by 2 to

fit

2 log3(x) = 8

log3(x) = 4

enb log

Switch

2 log3(x) = 8

log3(x) = 4

nben e

b log

Switch

2 log3(x) = 8

log3(x) = 4

34=x

nben eb log

Then Simplify

2 log3(x) = 8

log3(x) = 4

34=x

x = 81

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

Need to use a log law

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

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

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

NMMN logloglog

Switch

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

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

nben e

b log

Switch

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

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

23=(x-1)(x+1)

nben e

b log

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!

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)

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

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

23=(x-1)(x+1) = x2 -1

x = +3 or -3

There’s more than one way to do this

13ln x

389.4

3

3

2)3ln(

1)3ln(2

1

1)3ln(

13ln

2

2

2

1

x

xe

xe

x

x

x

x

Can you find why each step is valid?

389.4

3

3

2)3ln(

1)3ln(2

1

1)3ln(

13ln

2

2

2

1

x

xe

xe

x

x

x

x

rules of exponents

multiply both sides by 2

- 3 to get exact answer

Approximate answer

MrM ar

a loglog

ennb be log

Here’s another way to solve the same equation.

389.4

3

3

3

3

13ln

2

2

22

1

x

xe

xe

xe

xe

x

exclude 2nd result

389.43

33

33

3

3

3

13ln

2

22

22

2

22

1

ex

xeorxe

xeorxe

xe

xe

xe

x

Square both sides

Simplify

52x - 5x – 12 = 0

Factor it. Think of

y2 - y-12=0

52x - 5x – 12 = 0

(5x – 4)(5x + 3) = 0

Set each factor = 0

52x - 5x – 12 = 0

(5x – 4)(5x + 3) = 0

5x – 4 = 0 or 5x + 3 = 0

Solve first factor’s equation

Solve 5x – 4 = 0

5x = 4

log54 = x

Solve other factor’s

equation

Solve 5x + 3 = 0

5x = -3

log5(-3) = x

Oops, we cannot have a

negative argument

Solve 5x + 3 = 0

5x = -3

log5(-3) = x

Exclude this solution.

Only the other factor’s solution

works

Solve 5x + 3 = 0

5x = -3

log5(-3) = x

2

)5ln(x

4x+2 = 5x

If M = N then

ln M = ln N

2

)5ln(x

4x+2 = 5x

2

)5ln(x

If M = N then ln M = ln N

4x+2 = 5x

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

4x+2 = 5x

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

MrM ar

a loglog

4x+2 = 5x

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

MrM ar

a loglog

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)

Get x terms on one side

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)

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)

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)

)5ln()4ln(

)4ln(2

x