Solving Exponential and Logarithmic Equations

As you know, algebra often requires you to solve equations to find unknown values. This is also true for exponential and logarithmic equations. There are some strategies that you can use, along with some properties you’ve learned, that you can use to solve those equations.

You may be able to look at an equation like `4^x=16` and solve it by asking yourself, “`4` to what power is `16`? `4^2` is `16`, so `x = 2`.” Equations like `4^x=17` are a bit more difficult. You know `x` must be a little more than `2`, because `17` is just a little more than `16`. One way to find `x` with more precision, though, is by using logarithms A calculation in which the exponent `y` in `x = b^y` is found when given `x` and `b`; the corresponding notation is `log_bx = y`. .

When you have solved other algebraic equations, you often relied on the idea that you can change both sides of the equation in the same way and still get a true equation. This is true with logarithms, too: If `x = y`, then `log_bx=log_by`, no matter what `b` is.

Let’s look at this with an equation whose solution you already know: `4^x=16`. You can use either the common logA logarithm using `10` as the base (`log_10`). or the natural logA logarithm using `e` as the base, written as `log_e`. . In the following example, you will use the common log.

Example
Problem	Solve `4^x=16`.
	`4^x=16` `log4^x=log16`	Take the common logarithm of both sides. (Remember, when no base is written, that means the base is `10`.) What can you do with that new equation?
	`log4^x=log16` `x log 4 = log 16`	Use the power property of logarithms to simplify the logarithm on the left side of the equation.
	`x log 4 = log 16` `x=log16/log4`	Remember that `log4` is a number. You can divide both sides of the equation by `log4` to get `x` by itself.
Answer	`x=(1.204...)/(0.602...)=2`	Use a calculator to evaluate the logarithms and the quotient.

Just as you knew, `x = 2`. Now let’s try it with our more difficult example, `4^x=17`. The procedure is exactly the same.

Example
Problem	Solve `4^x=17`.
	`4^x=17` `log4^x=log17`	Take the common logarithm of both sides.
	`log4^x=log17` `xlog4 = "log"17`	Use the power property of logarithms to simplify the logarithm on the left.
	`xlog4 = "log"17` `x=log17/log4`	Divide both sides by `log4` to get `x` by itself.
Answer	`x=(1.230...)/(0.602...)=2.043...`	Use a calculator to evaluate the logarithms and the quotient.

You could have used either the common log or the natural log with the example above. You use one of these two bases, as you can then use your calculator to find the values.

Example
Problem	Solve `e^(2x)=54`.
	`e^(2x)=54` `lne^(2x)=ln54`	Since the base is `e`, use the natural logarithm. (If the base were `10`, using common logarithms would be better.)
	`lne^(2x)=ln54` `2x = ln54`	Remember that logarithms and exponential functions are inverses. When you have `log_b b^m`, the logarithm undoes the exponent, and the result is just `m`. So `lne^(2x)=log_e e^(2x)=2x`.
	`x=ln54/2`	Divide both sides by `2` to get `x` by itself.
Answer	`x=1.99449...`	Use a calculator to evaluate the logarithm and quotient on the right and you’re done!

Another kind of exponential equation has exponential expressions on both sides. When the bases are the same, or the exponents are the same, you can just compare the parts that are different. Look at these examples.

Example
Problem	Solve `3^(2x+5)=3^(3x-2)`.
	`3^(2x+5)=3^(3x-2)`	Here are two exponential expressions with the same base. If the two expressions are equal, then their exponents must be equal. (Think about that: if you have `3^a` and `3^b`, and `a!=b`, then `3^a` can’t have the same value as `3^b`.)
	`2x + 5 = 3x - 2`	So, write a new equation that sets the exponents equal to each other.
	`5 = x - 2` `7 = x`	Solve the linear equation as you normally would.
*Check*	`3^(2(7)+5)=3^(3(7)-2)` `3^19 = 3^19`	Test the solution in the original equation. No need to find `3^19`. When both sides say the same thing, you know it’s correct!
Answer	`x=7`

Example
Problem	Solve `(x + 4)^8 = 7^8`.
	`(x + 4)^8 = 7^8`	Again, you have two exponential expressions that are equal to each other. In this case, both sides have the same exponent, and this means the bases must be equal.
	`x + 4 = 7`	Write a new equation that sets the bases equal to each other.
	`x = 3`	Solve the linear equation as you normally would.
*Check*	`(3 + 4)^8 = 7^8` `7^8 = 7^8`	Test the solution in the original equation. No need to find `7^8`. When both sides say the same thing, you know it’s correct!
Answer	`x=3`

There are several strategies you can use to solve logarithmic equations. The first is one you have used before: Rewrite the logarithmic equation as an exponential equation!

Example
Problem	Solve `ln x = 4.657`. Give `x` to the thousandths place.
	`ln x = 4.657` `log_e x = 4.657` `e^4.657=x`	Remember that natural logarithms have a base of `e`. Rewrite this logarithm as an exponential equation.
Answer	`105.3196...=x` `x~~105.320`	Use a calculator to evaluate `e^4.657`, and round to the nearest thousandth.

Example
Problem	Solve `log_7x = 3.843`. Give `x` to the thousandths place.
	`log_7x = 3.843` `7^3.843=x`	Rewrite this logarithm as an exponential equation.
Answer	`1768.9345...=x` `x~~1768.935`	Use a calculator to evaluate `7^3.843` and round to the nearest thousandth.

Logarithmic equations may also involve inputs where the variable has a coefficient other than `1`, or where the variable itself is squared. In these cases, you need to complete a few more steps in solving for the variable.

Example
Problem	Solve `log_5 \3x^2 = 1.96`. Give `x` to the hundredths place.
	`5^1.96=3x^2`	Rewrite this logarithmic equation as an exponential equation.
	`23.44127...=3x^2`	Evaluate `5^1.96`.
	`7.81375...=x^2` `x=+-2.7953...` `x~~+-2.80`	Solve as you normally would. In this case, divide both sides by `3`, then use the square root property to find the possible values for `x`. Don’t forget that when using the square root property, both positive and negative roots must be considered. Round to the nearest hundredth.
*Check*	`log_5\3x^2 = 1.96` `log_5\3(-2.80)^2 = 1.96` `log_5\3(7.84) = 1.96` `log_5\23.52 = 1.96`	Check your answer by substituting the value of `x` into the original equation. Because `(-2.80)^2` and `(+2.80)^2` are both positive, we don't need to check `+\ 2.80` separately.
	`log23.52/log5=1.96` `(1.3714...)/(0.6989...)=1.96` `1.96=1.96`	Apply the change of base formula to switch from base `5` to base `10`. The check shows that with rounding accounted for, a true statement results, so you know that the answer is correct.
Answer	`x~~+-\ 2.80`

The equations may also include more than one logarithm. You can use the properties of logarithms to combine these logarithms into one logarithm. Note: You’ll find it helpful to record which properties you use at each step, both to help you be sure you’re using them properly and as a way to help you find errors.

Example
Problem		Solve `2log3+1/2log16-log3=logx`.
	`2log3+1/2log16-log3=logx`		First notice that all of the logarithms have the same base. (These are common logarithms, so the bases are all `10`.) When using the properties, it is absolutely necessary that the bases are the same.
	`log3^2+log16^(1/2)-log3=logx`		Use the power property to rewrite `2 log3` as `log3^2` and `1/2log16` as `log16^(1/2)`.
	`log9+log4-log3 = logx`		Evaluate the exponents.
	`log(9 * 4) -log3 = logx` `log36 -log3 = logx`		Use the product property `log_bMN=log_bM+log_bN` to combine `log9 +log4`.
	`log(36/3)=logx` `log12 = logx`		Use the quotient property `log_b(M/N)=log_bM-log_bN` to combine `log36 - log3`.
Answer	`x=12`		Since the logarithm of `12` and the logarithm of `x` are equal, `x` must equal `12`.

There are several strategies that can be used to solve equations involving exponents and logarithms. Taking logarithms of both sides is helpful with exponential equations. Rewriting a logarithmic equation as an exponential equation is a useful strategy. Using properties of logarithms is helpful to combine many logarithms into a single one.

Learning Objectives

Introduction

Solving Exponential Equations

Solving Logarithmic Equations

Summary