7.6A
Solving Exponential and
Logarithmic Equations
Algebra II
Exponential Equations
• One way to solve exponential
equations is to use the property that if
2 powers w/ the same base are equal,
then their exponents are equal.
• For b>0 & b≠1 if
x
b
=
y
b,
then x=y
Ex. 1)Solve by equating
exponents
3x
•4
x+1
8
=
2
3x
3
x+1
• (2 ) = (2 ) rewrite w/ same base
6x
3x+3
•2 = 2
Check
→
4
=
8
• 6x = 3x+3
64 = 64
•x = 1
3*1
1+1
Ex. 2) Solve by equating
exponents
4x
•2
x-1
32
=
4x
5
x-1
• 2 = (2 )
• 4x = 5x-5
•5 = x
Be sure to check your answer!!!
Ex. 3 When you can’t
rewrite using the same
base, you can solve by
taking a log of both sides
• 2x = 7
x
• log22 = log27
• x = log27
log 7
•x=
≈ 2.807
log 2
Ex. 4) Solve by equating exponents
Solve 4  15
x
• log4 = log415
• x = log415 = log15/log4
• ≈ 1.953
x
4
Ex. 5 5x+2 + 3 = 25
•
•
•
•
•
•
x+2
5
= 22
x+2
log55 = log522
x+2 = log522
x = (log522) – 2
= (log22/log5) – 2
≈ -.079
Ex.6
•
•
•
•
•
•
•
2x-3
10 +4
=
21
-4
-4
102x-3 = 17
log10102x-3 = log1017
2x-3 = log 17
2x = 3 + log17
x = ½(3 + log17)
≈ 2.115
Assignment
7.6B Solving Log
Equations
• To solve use the property for logs w/
the same base:
• + #’s b,x,y & b≠1
• If logbx = logby, then x = y
Ex. 1
log3(5x-1) = log3(x+7)
• 5x – 1 = x + 7
•
5x = x + 8
•
4x = 8
•
x = 2 and check
• log3(5*2-1) = log3(2+7)
•
log39 = log39
When faced with log/logs
on one side of equation,
then exponentiate each
side.
• b>0 & b≠1
•if x = y, then
x
b
=
y
b
Ex. 2)
•
•
•
log5(3x + 1) = 2
log
(3x+1)
5 5
=
2
5
3x+1 = 25
x = 8 and check
• Because the domain of log functions doesn’t include all
reals, you should check for extraneous solutions
Ex. 3 log2x + log2(x-7) = 3
•
•
•
•
•
•
•
log2x(x-7) = 3
log2 (x2- 7x) = 3
3
log
x²
-7x
2 2
=2
x2 – 7x = 8
x2 – 7x – 8 = 0
(x-8)(x+1)=0
x=8 x= -1
Ex. 4
log5x + log(x-1)=2
• log (5x)(x-1) = 2
(product property)
• log (5x2 – 5x) = 2
•
•
•
2
log5x
-5x
10
=
2
10
5x2 - 5x = 100
x2 – x - 20 = 0
(subtract 100 and divide by 5)
•
(x-5)(x+4) = 0 x=5, x=-4
• graph and you’ll see 5=x is the only solution
EX. 5
Newton’s Law of Cooling
• The temperature T of a cooling substance
@ time t (in minutes) is:
•T = (T0 – TR)
-rt
e
+ TR
• T0= initial temperature
• TR= room temperature
• r = constant cooling rate of the substance
• You’re cooking stew. When
you take it off the stove the
temp. is 212°F. The room
temp. is 70°F and the cooling
rate of the stew is r =.046. How
long will it take to cool the stew
to a serving temp. of 100°?
• T0 = 212, TR = 70, T = 100 r = .046
• So solve:
-.046t
• 100 = (212 – 70)e
+70
-.046t
• 30 = 142e
(subtract 70)
-.046t
• .221 ≈ e
(divide by 142)
• How do you get the variable out of the
exponent?
Cooling cont.
-.046t
e
• ln .221 ≈ ln
(take the ln of both sides)
• ln .221 ≈ -.046t
• -1.556 ≈ -.046t
• 33.8 ≈ t
• about 34 minutes to cool!