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When does an extraneous solution occur? How can an extraneous solution be recognized?

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When can the one-to-one property of logarithms be used to solve an equation? When can it not be used?

The one-to-one property can be used if both sides of the equation can be rewritten as a single logarithm with the same base. If so, the arguments can be set equal to each other, and the resulting equation can be solved algebraically. The one-to-one property cannot be used when each side of the equation cannot be rewritten as a single logarithm with the same base.

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Algebraic

For the following exercises, use like bases to solve the exponential equation.

4 3 v 2 = 4 v

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64 4 3 x = 16

x = 1 3

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2 3 n 1 4 = 2 n + 2

n = 1

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36 3 b 36 2 b = 216 2 b

b = 6 5

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( 1 64 ) 3 n 8 = 2 6

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For the following exercises, use logarithms to solve.

e r + 10 10 = −42

No solution

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8 10 p + 7 7 = −24

p = log ( 17 8 ) 7

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7 e 3 n 5 + 5 = −89

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e 3 k + 6 = 44

k = ln ( 38 ) 3

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5 e 9 x 8 8 = −62

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6 e 9 x + 8 + 2 = −74

x = ln ( 38 3 ) 8 9

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e 2 x e x 132 = 0

x = ln 12  

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7 e 8 x + 8 5 = −95

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10 e 8 x + 3 + 2 = 8

x = ln ( 3 5 ) 3 8

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8 e 5 x 2 4 = −90

no solution

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e 2 x e x 6 = 0

x = ln ( 3 )

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3 e 3 3 x + 6 = −31

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For the following exercises, use the definition of a logarithm to rewrite the equation as an exponential equation.

log ( 1 100 ) = −2

10 2 = 1 100

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For the following exercises, use the definition of a logarithm to solve the equation.

4 + log 2 ( 9 k ) = 2

k = 1 36

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2 log ( 8 n + 4 ) + 6 = 10

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10 4 ln ( 9 8 x ) = 6

x = 9 e 8

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For the following exercises, use the one-to-one property of logarithms to solve.

ln ( 10 3 x ) = ln ( 4 x )

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log 13 ( 5 n 2 ) = log 13 ( 8 5 n )

n = 1

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log ( x + 3 ) log ( x ) = log ( 74 )

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ln ( 3 x ) = ln ( x 2 6 x )

No solution

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log 4 ( 6 m ) = log 4 3 m

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ln ( x 2 ) ln ( x ) = ln ( 54 )

No solution

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log 9 ( 2 n 2 14 n ) = log 9 ( 45 + n 2 )

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ln ( x 2 10 ) + ln ( 9 ) = ln ( 10 )

x = ± 10 3

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For the following exercises, solve each equation for x .

log ( x + 12 ) = log ( x ) + log ( 12 )

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ln ( x ) + ln ( x 3 ) = ln ( 7 x )

x = 10

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ln ( 7 ) + ln ( 2 4 x 2 ) = ln ( 14 )

x = 0

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log 8 ( x + 6 ) log 8 ( x ) = log 8 ( 58 )

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ln ( 3 ) ln ( 3 3 x ) = ln ( 4 )

x = 3 4

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log 3 ( 3 x ) log 3 ( 6 ) = log 3 ( 77 )

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Graphical

For the following exercises, solve the equation for x , if there is a solution . Then graph both sides of the equation, and observe the point of intersection (if it exists) to verify the solution.

log 9 ( x ) 5 = −4

x = 9

Graph of log_9(x)-5=y and y=-4.
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ln ( 3 x ) = 2

x = e 2 3 2.5

Graph of ln(3x)=y and y=2.
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log ( 4 ) + log ( 5 x ) = 2

x = 5

Graph of log(4)+log(-5x)=y and y=2.
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7 + log 3 ( 4 x ) = −6

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ln ( 4 x 10 ) 6 = 5

x = e + 10 4 3.2

Graph of ln(4x-10)-6=y and y=-5.
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log ( 4 2 x ) = log ( 4 x )

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log 11 ( 2 x 2 7 x ) = log 11 ( x 2 )

No solution

Graph of log_11(-2x^2-7x)=y and y=log_11(x-2).
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ln ( 2 x + 9 ) = ln ( 5 x )

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log 9 ( 3 x ) = log 9 ( 4 x 8 )

x = 11 5 2.2

Graph of log_9(3-x)=y and y=log_9(4x-8).
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log ( x 2 + 13 ) = log ( 7 x + 3 )

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3 log 2 ( 10 ) log ( x 9 ) = log ( 44 )

x = 101 11 9.2

Graph of 3/log_2(10)-log(x-9)=y and y=log(44).
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ln ( x ) ln ( x + 3 ) = ln ( 6 )

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For the following exercises, solve for the indicated value, and graph the situation showing the solution point.

An account with an initial deposit of $6,500 earns 7.25 % annual interest, compounded continuously. How much will the account be worth after 20 years?

about $ 27 , 710.24

Graph of f(x)=6500e^(0.0725x) with the labeled point at (20, 27710.24).
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The formula for measuring sound intensity in decibels D is defined by the equation D = 10 log ( I I 0 ) , where I is the intensity of the sound in watts per square meter and I 0 = 10 12 is the lowest level of sound that the average person can hear. How many decibels are emitted from a jet plane with a sound intensity of 8.3 10 2 watts per square meter?

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The population of a small town is modeled by the equation P = 1650 e 0.5 t where t is measured in years. In approximately how many years will the town’s population reach 20,000?

about 5 years

Graph of P(t)=1650e^(0.5x) with the labeled point at (5, 20000).
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Technology

For the following exercises, solve each equation by rewriting the exponential expression using the indicated logarithm. Then use a calculator to approximate x to 3 decimal places .

1000 ( 1.03 ) t = 5000 using the common log.

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e 5 x = 17 using the natural log

ln ( 17 ) 5 0.567

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3 ( 1.04 ) 3 t = 8 using the common log

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3 4 x 5 = 38 using the common log

x = log ( 38 ) + 5 log ( 3 )     4 log ( 3 ) 2.078

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50 e 0.12 t = 10 using the natural log

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For the following exercises, use a calculator to solve the equation. Unless indicated otherwise, round all answers to the nearest ten-thousandth.

7 e 3 x 5 + 7.9 = 47

x 2.2401

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ln ( 3 ) + ln ( 4.4 x + 6.8 ) = 2

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log ( 0.7 x 9 ) = 1 + 5 log ( 5 )

x 44655 . 7143

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Atmospheric pressure P in pounds per square inch is represented by the formula P = 14.7 e 0.21 x , where x is the number of miles above sea level. To the nearest foot, how high is the peak of a mountain with an atmospheric pressure of 8.369 pounds per square inch? ( Hint : there are 5280 feet in a mile)

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The magnitude M of an earthquake is represented by the equation M = 2 3 log ( E E 0 ) where E is the amount of energy released by the earthquake in joules and E 0 = 10 4.4 is the assigned minimal measure released by an earthquake. To the nearest hundredth, what would the magnitude be of an earthquake releasing 1.4 10 13 joules of energy?

about 5.83

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Extensions

Use the definition of a logarithm along with the one-to-one property of logarithms to prove that b log b x = x .

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Recall the formula for continually compounding interest, y = A e k t . Use the definition of a logarithm along with properties of logarithms to solve the formula for time t such that t is equal to a single logarithm.

t = ln ( ( y A ) 1 k )

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Recall the compound interest formula A = a ( 1 + r k ) k t . Use the definition of a logarithm along with properties of logarithms to solve the formula for time t .

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Newton’s Law of Cooling states that the temperature T of an object at any time t can be described by the equation T = T s + ( T 0 T s ) e k t , where T s is the temperature of the surrounding environment, T 0 is the initial temperature of the object, and k is the cooling rate. Use the definition of a logarithm along with properties of logarithms to solve the formula for time t such that t is equal to a single logarithm.

t = ln ( ( T T s T 0 T s ) 1 k )

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Questions & Answers

what is subgroup
Purshotam Reply
Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
Macmillan Reply
e power cos hyperbolic (x+iy)
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10y
Michael
tan hyperbolic inverse (x+iy)=alpha +i bita
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prove that cos(π/6-a)*cos(π/3+b)-sin(π/6-a)*sin(π/3+b)=sin(a-b)
Tejas Reply
why {2kπ} union {kπ}={kπ}?
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if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41
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-4
Joel
x=-4
Joel
x=-1
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x=-1
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Adityasuman x= - 1
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-1
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Practice Key Terms 1

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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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