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In this section, you will:
  • Use like bases to solve exponential equations.
  • Use logarithms to solve exponential equations.
  • Use the definition of a logarithm to solve logarithmic equations.
  • Use the one-to-one property of logarithms to solve logarithmic equations.
  • Solve applied problems involving exponential and logarithmic equations.
Seven rabbits in front of a brick building.
Wild rabbits in Australia. The rabbit population grew so quickly in Australia that the event became known as the “rabbit plague.” (credit: Richard Taylor, Flickr)

In 1859, an Australian landowner named Thomas Austin released 24 rabbits into the wild for hunting. Because Australia had few predators and ample food, the rabbit population exploded. In fewer than ten years, the rabbit population numbered in the millions.

Uncontrolled population growth, as in the wild rabbits in Australia, can be modeled with exponential functions. Equations resulting from those exponential functions can be solved to analyze and make predictions about exponential growth. In this section, we will learn techniques for solving exponential functions.

Using like bases to solve exponential equations

The first technique involves two functions with like bases. Recall that the one-to-one property of exponential functions tells us that, for any real numbers b , S , and T , where b > 0 ,   b 1 , b S = b T if and only if S = T .

In other words, when an exponential equation has the same base on each side, the exponents must be equal. This also applies when the exponents are algebraic expressions. Therefore, we can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then, we use the fact that exponential functions are one-to-one to set the exponents equal to one another, and solve for the unknown.

For example, consider the equation 3 4 x 7 = 3 2 x 3 . To solve for x , we use the division property of exponents to rewrite the right side so that both sides have the common base, 3. Then we apply the one-to-one property of exponents by setting the exponents equal to one another and solving for x :

3 4 x 7 = 3 2 x 3 3 4 x 7 = 3 2 x 3 1 Rewrite 3 as 3 1 . 3 4 x 7 = 3 2 x 1 Use the division property of exponents . 4 x 7 = 2 x 1   Apply the one-to-one property of exponents . 2 x = 6 Subtract 2 x  and add 7 to both sides . x = 3 Divide by 3 .

Using the one-to-one property of exponential functions to solve exponential equations

For any algebraic expressions S  and  T , and any positive real number b 1 ,

b S = b T if and only if S = T

Given an exponential equation with the form b S = b T , where S and T are algebraic expressions with an unknown, solve for the unknown.

  1. Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form b S = b T .
  2. Use the one-to-one property to set the exponents equal.
  3. Solve the resulting equation, S = T , for the unknown.

Solving an exponential equation with a common base

Solve 2 x 1 = 2 2 x 4 .

  2 x 1 = 2 2 x 4 The common base is   2.      x 1 = 2 x 4 By the one-to-one property the exponents must be equal .           x = 3 Solve for  x .
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Solve 5 2 x = 5 3 x + 2 .

x = 2

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Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
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can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
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Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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