Using the Definition of a Logarithm to Solve Logarithmic Equations

Site:	Saylor Academy
Course:	MA001: College Algebra
Book:	Using the Definition of a Logarithm to Solve Logarithmic Equations

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Date:	Thursday, 3 April 2025, 12:03 AM

Description

In this unit, you will explore the techniques for solving logarithmic equations. We will begin by using the definition of a logarithm to "undo" it. Then, we will work up to more complex techniques.

Using the Definition of a Logarithm to Solve Logarithmic Equations
Using the One-to-One Property of Logarithms to Solve Logarithmic Equations
Solve log equations using the definition of logarithm
Solving Simple Log equations

Using the Definition of a Logarithm to Solve Logarithmic Equations

We have already seen that every logarithmic equation $log_b(x)=y$ is equivalent to the exponential equation $b^y=x$ . We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression.

For example, consider the equation $log_2(2)+log_2(3x−5)=3$ . To solve this equation, we can use rules of logarithms to rewrite the left side in compact form and then apply the definition of logs to solve for 5 $x$ :

$log_2(2)+log_2(3x−5)=3$
$log_2(2(3x−5))=3$	Apply the product rule of logarithms.
$log_2(6x−10)=3$	Distribute.
$2^3=6x−10$	Apply the definition of a logarithm.
$8=6x−10$	Calculate $2^3$ .
$18=6x$	Add 10 to both sides.
$x=3$	Divide by 6.

Using the Definition of a Logarithm to Solve Logarithmic Equations

For any algebraic expression $S$ and real numbers $b$ and $c$ , where $b > 0$ , $b≠1$ ,

$log_b(S)=c$ if and only if $b^c=S$

Example 9

Using Algebra to Solve a Logarithmic Equation

Solve $2lnx+3=7$ .

Solution

$2lnx+3=7$
$2lnx=4$	Subtract 3.
$lnx=2$	Divide by 2.
$x=e^2$	Rewrite in exponential form.

Try It #9

Solve $6+lnx=10$ .

Example 10

Using Algebra Before and After Using the Definition of the Natural Logarithm

Solve $2ln(6x)=7$ .

Solution

$2ln(6x)=7$
$In (6x) = \dfrac{7}{2}$	Divide by 2.
$6x = e^{(\dfrac{7}{2})}$	Use the definition of $ln$ .
$x = \dfrac{1}{6} e^{(\dfrac{7}{2})}$	Divide by 6.

Try It #10

Solve $2ln(x+1)=10$ .

Example 11

Using a Graph to Understand the Solution to a Logarithmic Equation

Solve $ln x=3$ .

Solution

$ln x=3$

$x=e^3$ Use the definition of the natural logarithm.

Figure 3 represents the graph of the equation. On the graph, the x-coordinate of the point at which the two graphs intersect is close to 20. In other words $e^3≈20$ . A calculator gives a better approximation: $e^3≈20.0855$ .

Graph of two questions, y=3 and y=ln(x), which intersect at the point (e^3, 3) which is approximately (20.0855, 3).

Figure 3 The graphs of $y=ln x$ and $y=3$ cross at the point $(e^3,3)$ , which is approximately (20.0855, 3).

Try It #11

Use a graphing calculator to estimate the approximate solution to the logarithmic equation $2^x=1000$ to 2 decimal places.

Source: Rice University, https://openstax.org/books/college-algebra/pages/6-6-exponential-and-logarithmic-equations
This work is licensed under a Creative Commons Attribution 4.0 License.

Using the One-to-One Property of Logarithms to Solve Logarithmic Equations

As with exponential equations, we can use the one-to-one property to solve logarithmic equations. The one-to-one property of logarithmic functions tells us that, for any real numbers

$x > 0$ ,

$S > 0$ ,

$T > 0$ and any positive real number

$b$ , where

$b≠1$ ,

$log_bS=log_bT$ if and only if

$S=T$ .

For example,

$log_2(x−1)=log_2(8)$ , then

$x−1=8$ .

So, if

$x−1=8$ , then we can solve for

$x$ , and we get

$x=9$ . To check, we can substitute

$x=9$ into the original equation:

$log_2(9−1)=log_2(8)=3$ . In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown.

For example, consider the equation

$log(3x−2)−log(2)=log(x+4)$ . To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm, and then apply the one-to-one property to solve for

$x$ :

$log(3x−2)−log(2)=log(x+4)$
$log(\dfrac{3x−2}{2})=log(x+4)$	Apply the quotient rule of logarithms.
$\dfrac{3x−2}{2}=x+4$	Apply the one to one property of a logarithm.
$3x−2=2x+8$	Multiply both sides of the equation by 2.
$x=10$	Subtract $2x$ and add 2.

To check the result, substitute

$x=10$ into

$log(3x−2)−log(2)=log(x+4)$ .

$log(3(10)−2)−log(2)=log((10)+4)$

$log(28)−log(2)=log(14)$

$log(\dfrac{28}{2})=log(14)$ The solution checks.

Using the One-to-One Property of Logarithms to Solve Logarithmic Equations

For any algebraic expressions

$S$ and

$T$ and any positive real number

$b$ , where

$b≠1$ ,

$log_bS=log_bT$ if and only if

$S=T$

Note, when solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution.

How To

Given an equation containing logarithms, solve it using the one-to-one property.

Use the rules of logarithms to combine like terms, if necessary, so that the resulting equation has the form $log_bS=log_bT$ .
Use the one-to-one property to set the arguments equal.
Solve the resulting equation, $S=T$ , for the unknown.

Example 12

Solving an Equation Using the One-to-One Property of Logarithms

Solve

$ln(x^2)=ln(2x+3)$ .

Solution

$ln(x^2)=ln(2x+3)$
$x^2=2x+3$	Use the one-to-one property of the logarithm.
$x^2−2x−3=0$	Get zero on one side before factoring.
$(x−3)(x+1)=0$	Factor using FOIL.
$x−3=0$ or $x+1=0$	If a product is zero, one of the factors must be zero.
$x=3$ or $x=−1$	Solve for $x$ .

Analysis

There are two solutions: 3 or −1. The solution −1 is negative, but it checks when substituted into the original equation because the argument of the logarithm functions is still positive.