3.4
Exponential and
Logarithmic Equations
Properties of Exp. and Log Functions
loga
ax
ln
ex
=x
=x
a
log a x
e
ln x
 x
 x
Solving Exponential Equations
Ex.
Take the ln of both sides.
ex = 72
ln ex = ln 72
x = ln 72  4 . 277
Ex.
4e2x = 5
e
2x

2 x  ln
4
5
x
4
ln e
2x
 ln
5
5
4
1
2
ln
5
4
 . 112
Solving an Exponential Equation
Ex.
2(32t-5) - 4 = 11
First, add 4 to each side
2(32t-5) = 15
Divide by 2
(32t-5) = 15/2
ln(32t-5) = ln 7.5
(2t-5) ln3 = ln 7.5
2tln3 - 5ln3 = ln 7.5
2tln3 = 5ln3 + ln 7.5
t
5 ln
3  ln 7 .5 
2 ln 3 
= 3.417
Ex.
e2x – 3ex + 2 = 0
(ex)2 – 3ex + 2 = 0
This factors.
( ex – 2 ) ( ex - 1 ) = 0
ex = 2
ln ex = ln 2
x = ln 2
ex = 1
ln ex = ln 1
x=0
Set both = 0 and finish
solving.
2x = 10
Ex.
ln 2x = ln 10
x
ln 10
 3 . 322
ln 2
x ln 2 = ln 10
Ex.
4x+3 = 7x
ln 4x+3 = ln 7x
x 
3 ln 4
ln 7  ln 4
 7 . 432
(x + 3) ln 4 = x ln 7
x ln 4 + 3 ln 4 = x ln 7
Collect like terms
3 ln 4 = x ln 7 - x ln 4
Factor out an x

3 ln 4 = x( ln 7 – ln 4)
Solving a Logarithmic Equation
Ex.
ln x = 2
Take both sides to the e
eln x = e2
x = e2  7 . 389
Ex.
5 + 2 ln x = 4
xe
2 ln x = -1
ln x = 
Ex.
1
3x = e2
2
 . 607
2
2 ln 3x = 4
ln 3x = 2
1
x
e
2
3
 2 . 463
Ex.
ln (x – 2) + ln (2x – 3) = 2 ln x
ln (x – 2)(2x – 3) = ln x2
2x2 – 7x + 6 = x2
+ means mult.
e to both sides
to get rid of ln’s.
x2 – 7x + 6 = 0
( x – 6 ) ( x – 1) = 0
6 and 1 are possible answers
Remember, can not take the log of a neg. number
or zero. Put answers back into the original to
check them.
Notice that only 6 works!