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Pre-Calc
Solving Logarithmic Equations
0
1. Isolate the logarithmic expression (logb y)
Lesson 5.6 C
Objective: SSBAT solve logarithmic equations.
Standards: 2.1.11F, 2.8.11B,D,
2.11.11B
2. Change to exponential form
3. Solve for the variable
4. Check for extraneous solutions
Examples: Solve each.
2.
1.
log x = 5
ln x = 5
105 = x
e5 = x
100000 = x
148.4132 = x
3.
log6 x = 4
64 = x
1296 = x
4.
5 + 2 ln x = 4
2 ln x = -1
ln x = -0.5
e-0.5 = x
0.6065 = x
1
.
5.
4 = 12
10 + log5
6.
2 log x = 3 log 4
4 = 2
log x2 = log 43
52 =
4
x2 = 64
25 =
4
x = 8
log5
252 = x – 4
625 = x – 4
629 = x
7.
ln x + ln (3x – 5) = ln 2
7. Continued
ln (x(3x – 5)) = ln 2
3x2 – 5x – 2 = 0
ln (3x2 – 5x) = ln 2
(3x + 1)(x – 2) = 0
(3x2 – 5x) = 2
3x + 1 = 0
and
x–2=0
x = -1/3
and
x= 2
3x2 – 5x – 2 = 0
 Continue on next slide
extraneous
Solution: x = 2
8.
ln (x – 3) + ln (2x + 1) = 2 ln x
8. Continued
ln ((x – 3)(2x + 1)) = ln x2
x=
(x – 3)(2x + 1) = x2
x=
∙
2x2 + x – 6x – 3 = x2
2x2 – 5x – 3 = x2
x2 – 5x – 3 = 0
 solve using quadratic formula – continue next slide
x=
x ≈ 5.5414
and
x=
x ≈ -0.5414
Extraneous
Answer: x = 5.5414
2
.
9. log x + log (x + 3) = 1
9. Continued
log (x(x + 3)) = 1
0 = x2 + 3x – 10
log (x2 + 3x) = 1
0 = (x – 2)(x + 5)
101 = x2 + 3x
x–2=0
10 = x2 + 3x
x=2
and x + 5 = 0
x = -5
Extraneous
0 = x2 + 3x – 10
Answer: x = 2
Solve by factoring
10.
ln x = ln 3 – ln (x + 5)
10.
Continued
x2 + 5x – 3 = 0
ln x = ln
x(x + 5) = 3
x=
x2 + 5x = 3
x2
x=
x =
and
x = 0.5414
+ 5x – 3 = 0
ln (3x – 5) = ln 11 + ln 2
ln (3x – 5) = 3.0910
x=
x = -5.5414
Extraneous
Solve by quadratic formula (next slide)
11.
Answer: x = 0.5414
12.
ln (x + 9) – ln x = 1
ln
= 1
e3.0910 = 3x – 5
e1 =
22 = 3x – 5
ex = x + 9
27 = 3x
ex – x = 9
9 = x
x(e – 1) = 9
x=
3