3.4
Exponential & Logarithmic Equations
JMerrill, 2010
Quick Review of 3.3
 Properties of Logs
Rules of Logarithms
If M and N are positive real numbers and b is ≠ 1:

The Product Rule:

logbMN = logbM + logbN
(The logarithm of a product is the sum of the logarithms)

Example: log (10x) = log10 + log x


You do: log7(1000x) =
log71000 + log7x
Rules of Logarithms
If M and N are positive real numbers and b ≠ 1:

The Quotient Rule
M 
logb    log b M  log b N
N
(The logarithm of a quotient is the difference of the logs)

x
log

log
x

log
2


Example:
2

 14 
You do: log 7  x 
 log7 14  log7 x
Rules of Logarithms
If M and N are positive real numbers, b ≠ 1, and p is
any real number:
 The Power Rule:
 logbMp = p logbM
(The log of a number with an exponent is the product of the
exponent and the log of that number)




Example: log x2 = 2 log x
Example: log574 = 4 log57
You do: log359 = 9 log35
1
Challenge: log x  log x 2  1 log x
2
Condensing
 Sometimes, we need to condense
before we can solve:
logb M  logb N  3logb P
logb MN  3log b P
logb MN  logb P
MN
logb 3
P
3
Product Rule
Power Rule
Quotient Rule
Condensing
 Condense:
1
 logb M  logb N  logb P 
2
 logb M  logb N  logb P 
log b
1
2
1
2
MN
 MN
or log b 

P
 P 
Using the Rules to Condense
 Ex:
1
ln20  2ln  ln x
2
2
1
 ln20  ln    ln x
2
1
 ln20   x  ln5x
4
 You Do:
2(ln2  ln x )  (ln x  ln 4)
 (ln2  ln x )2  ln x  ln 4
2
2
 ln    ln x  ln 4 
x 
4x
1
ln
 ln
2
x
4x
Bases
 We don’t really use other bases anymore,
but since logs are often written in other
bases, we must change to base 10 in order
to use our calculators.
Change of Base Formula
log(c)
log b c 
log(b)
log(8)
 1.290
 Example log58 =
log(5)
Parentheses are
vital! The log key
opens the ( ), you
must close it!
 This is also how you graph in another base.
Enter y1=log(8)/log(5). Remember, you don’t
have to enter the base when you’re in base 10!
3.4
Solving Exponents & Logs
Solving Guidelines
Get both parts to
the same base
 Original
Rewritten





2x = 25
x=5
lnx = ln3
x=3
3-x = 32
x = -2
lnex = ln7
x = ln7
10logx = 10-1
x = 10-1
= 1/10
If you have a log on
2x = 32
lnx – ln3 = 0
(1/3)x = 9
ex = 7
logx =-1
Solution
If you have a
Solvevariable
like
in the
Get
both partsposition,
to
normal
exponent
thetake
same
thebase
log of both
sides. Take the ln if
you’re using e, take
the log if using
common logs.
one side,
exponentiate both
sides
Solving
 Getting all the numbers to the same
base.
x 1
9
 27
 Example:
1
2 
8
1
x
2  3
2
x
3
2 2
x
x  3
9 x 1  27
3 
2 x 1
1
2
1
3 2
 3
32 x  2  3
3
2
3
2x  2 
2
1
x
4

Solving
 Clear the exponent:
3
2
4 x  32
3
2
x 8
2
3
2
 32 
3
x

8
 
 
x4
 x  1
 x  1
1
4
2 0
1
4
2
4


4
x

1

2






1
x 1 
16
17
x
16
1
4
Solving Exponentials




Exponentiating:
ex = 72
lnex = ln72
x = ln72 ≈ 4.277
 You should always check your answers by
plugging them back in. Sometimes they
don’t work because you can’t take the log of
a negative number.
Solving Exponentials





3(2x) = 42
AlternativeMethod :
2x = 14
3(2x )  42
log22x = log214
2x  14
x = log214
ln2x  ln14
x = log14/log2 ≈ 3.807 xln2  ln14
ln14
x
 3.807
ln2
Solving Exponentials






4e2x – 3 = 2
4e2x = 5
e2x = 5/4
lne2x = ln 5/4
2x = ln 5/4
x = ½ ln 5/4 ≈ 0.112
Solving Exponentials








2(32t-5) – 4 = 11
2(32t-5) = 15
(32t-5)= 15/2
log3(32t-5) = log3 15/2
2t – 5 = log3 15/2
2t = 5 + log3 7.5
t = 5/2 + ½ log3 7.5
t ≈ 3.417
Solving Exponentials








e2x – 3ex + 2 = 0
No like terms—kinda look quadratic?
(ex – 2)(ex – 1) = 0
Set each factor = 0 and solve
(ex – 2) = 0
(ex – 1) = 0
ex = 2
x = 1
e
lnex = ln2
x = ln 1
lne
x = ln2 ≈ 0.693
x=0
Solving Logarithms




Exponentiating with the natural log
lnx = 2
elnx = e2
x = e2 ≈ 7.389
Solving Logarithms




log3(5x - 1) = log3(x + 7)
5x – 1 = x + 7
4x = 8
x=2
Solving Logs – Last Time






5 + 2lnx = 4
2lnx = -1
lnx = - ½
elnx = e - ½
x = e -½
x ≈ 0.607
Interest Compounded Continuously
 If interest is compounded “all the time”
(MUST use the word continuously), we use
the formula
P(t )  P0 e
rt
where P0 is the initial principle (initial
amount)
P(t )  P0 e
rt
 If you invest $1.00 at a 7% annual rate that
is compounded continuously, how much will
you have in 4 years?
1* e
(.07)(4)
 1.3231
 You will have a whopping $1.32 in 4 years!