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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

preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
A hedge is contrusted to be in the shape of hyperbola near a fountain at the center of yard.the hedge will follow the asymptotes y=x and y=-x and closest distance near the distance to the centre fountain at 5 yards find the eqution of the hyperbola
ayesha Reply
A doctor prescribes 125 milligrams of a therapeutic drug that decays by about 30% each hour. To the nearest hour, what is the half-life of the drug?
Sandra Reply
Find the domain of the function in interval or inequality notation f(x)=4-9x+3x^2
prince Reply
Jessica Reply
Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the high temperature of ?105°F??105°F? occurs at 5PM and the average temperature for the day is ?85°F.??85°F.? Find the temperature, to the nearest degree, at 9AM.
Karlee Reply
if you have the amplitude and the period and the phase shift ho would you know where to start and where to end?
Jean Reply
rotation by 80 of (x^2/9)-(y^2/16)=1
Garrett Reply
thanks the domain is good but a i would like to get some other examples of how to find the range of a function
bashiir Reply
what is the standard form if the focus is at (0,2) ?
Lorejean Reply
Roy Reply
Roy Reply
Roy Reply
A bridge is to be built in the shape of a semi-elliptical arch and is to have a span of 120 feet. The height of the arch at a distance of 40 feet from the center is to be 8 feet. Find the height of the arch at its center
Abdulfatah Reply
Practice Key Terms 1

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