1. 5.6 - Solving Logarithmic and Exponential Equations
Exponentials:
ln(ex) = x…
so if ex = 2…
ln(ex) = ln(2)
x = ln(2)
ex:
ex + 5 = 60
ex = 55
ln(ex) = ln(55)
x = ln(55)
ex:
4e2x = 5
e2x = 5/4
ln(e2x) = ln(5/4)
2x = ln(5/4)
x = ln(5/4) / 2

2. ex:
e2x - 3ex + 2 = 0
(ex - 2) (ex - 1) = 0
ex: Other Bases
2x = 10
ln(2x) = ln(10)
x ln(2) = ln(10)
x = ln(10) / ln(2)
= 3.322
(ex - 2) = 0
ex = 2
ln(ex) = ln(2)
x = ln(2)
= 0.693
(ex - 1) = 0
ex = 1
ln(ex) = ln(1)
x = 0

3. Finding Solutions Graphically2)
Set equation = 0
Graph
Find zero(s)
ex: 2x = 10 3) 1)

4. IC 5.6A - pg 386
#’s 1, 2, 4, 5, 8, 10, 11, 14, 15, 18, 19, 22, 25, 26, 30, 31, 34.

5. e(ln 3x) = e2 Solving Logarithmic Equations
Exponentials and logs are inverse functions
So for logs we raise both sides of the equation up as exponents…the base of the exponential we are creating is the base of the log statement (e for ln’s / whatever ‘a’ is for logs)
We call this ‘exponentiation’
ex:
ln x = 3
eln x = e3
x = e3
5 + 2ln x = 4
2ln x = -1
ln x = -1/2
eln x = e(-1/2)
x = e(-1/2)
2ln 3x = 4
ln 3x = 2
e(ln 3x) = e2
3x = e2
x = e2/3

6. 2(log23x) = 22 = 64 Other Bases ex: log4x = 3 4(log4x) = 43 x = 43

7. Shortcut
ex:
log10(x + 2) = log10 (3x - 4)
Drop log statements
(x + 2) = (3x - 4)
2x = 6
x = 3
Multiple log (or ln) statements
ln(x - 3) + ln(2x + 1) = 2(ln x)
Combine with laws of logs
ln[(x – 3)(2x + 1)] = ln x2
Drop log statements
(x – 3)(2x + 1) = x2
2x2 - 5x – 3 = x2
x2 - 5x – 3 = 0
Factor / Graph

8. Creating log (or ln) statements from constants
loga ax = x so…
log10 10 = 1
ln e = 1
ln e2 = 2
ln e3 = 3
log = 2
log = 3
etc…
ex: 2 + ln (x + 4) = ln 16
ln e2 + ln (x + 4) = ln 16
ln e2 (x + 4) = ln 16
Combine log statements
e2 (x + 4) = 16
Drop log statements
x = (16/e2) - 4

9. IC 5.6B – pg 386
#’s 37, 38, 41, 42, all, 52.