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1. Use a property of logarithms to evaluate log392
2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150.
3. Expand ln 732x4.
4. Condense 2log5y – 2log55 + ½log5x.
5. Use the change of base formula to evaluate log820.
1Algebra II
Solving Exponential and Logarithmic Equations
Algebra II
To solve equations you must undo what is being done
Exponential and logarithmic functions are inverse operations
You undo “exponentiating” by ”logarizing” and you undo “logarizing” by “exponentiating”
3Algebra II
1. ax = ay If x = y2. Logbx = logby if x = yExamples:
If 2x = 23 then x = 3 If ln x = 3 then e lnx = e3, so x = e3 If log3x = log3(5), then x = 5 If ex = 7 then ln ex = ln7 , so x = ln7
4Algebra II
Algebra II 5
Algebra II 6
83 xx
82 x4x
1432 xx
2x
Algebra II 7
xx 314 1x12 x
8Algebra II
9Algebra II
2x
42 x
521 log01 log 2 x
152 log2 x2 2
152102 x
10Algebra II
1xe
0x
44 xe21ln ln 2 xe
12ln 2 x2 2
122 xe
11Algebra II
133 xx123 x
x22 1x
12Algebra II
121 xx11 x
2xNo Solution
(Extraneous solution)
13Algebra II
1144 xx1154 x
x515 3x
14Algebra II
9x
15Algebra II
23x
22 log5 x
16Algebra II
4
52 ln x
17Algebra II
Check both
solutions!
18Algebra II
19Algebra II
20Algebra II
21Algebra II
22Algebra II
1. ln x = x2 – 2 2. Log3x = x2 – 2
23Algebra II
The population size y of a community of lemmings varies according to the relationship y = y0e0.15t. In this formula, t
is time in months and y0 is the initial population at the time 0. Estimate the population after 8 months if there
were originally 2000 lemmings.
y = y0e0.15t
y = 2000e0.15(8)
y = 2000e1.2
y ≈ 6640.2339
In 8 months, the population will be approximately 6640 lemmings.
Algebra II 24
How long does it take an investment of $5,000 to double if it is invested at 4%, compounded
quarterly?P = $5000r = 4% or 0.04Compounded quarterly = 4 times per year, n = 4.The investment doubles, so A must be $10,000.
Substitute these values and solve for t.
Algebra II 25
It takes more than17 years for themoney to double
in value.
Algebra II 26
27Algebra II
Solve each equation. Be sure to check solutions!!
1.32x = 27x+2
2.5e3x + 2 = 17
3.log4(5x – 11) = log4(3 – 2x)
4.-3 ln = 4