Transcript 8.6 Solving Exponential and Logarithmic Equations p. 501 Exponential Equations • One way to solve exponential equations is to use the property that if 2 powers.

8.6
Solving Exponential and
Logarithmic Equations
p. 501
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
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
Your turn!
4x
•2
x-1
32
=
4x
5
x-1
• 2 = (2 )
• 4x = 5x-5
•5 = x
Be sure to check your answer!!!
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
x
4
= 15
• log4 = log415
• x = log415 = log15/log4
• ≈ 1.95
x
4
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
•
•
•
•
•
•
x+2
5
+ 3 = 25
x+2
5
= 22
x+2
log55 = log522
x+2 = log522
x = (log522) – 2
= (log22/log5) – 2
≈ -.079
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!
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
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 you can’t rewrite
both sides as logs w/ the
same base exponentiate
each side
• b>0 & b≠1
•if x = y, then
x
b
=
y
b
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
log5x + log(x+1)=2
• log (5x)(x+1) = 2
• log (5x2 – 5x) = 2
•
•
•
2
log5x
-5x
10
=
(product property)
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
One More!
log2x + log2(x-7) = 3
• log2x(x-7) = 3
• log2 (x2- 7x) = 3
• 2log2x -7x = 32
• x2 – 7x = 8
2
• x – 7x – 8 = 0
• (x-8)(x+1)=0
• x=8 x= -1
2
Assignment