We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

High School Math : Solving and Graphing Logarithmic Equations

Study concepts, example questions & explanations for High School Math

Create An Account

All High School Math Resources

8 Diagnostic Tests 613 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

← Previous 1 2 Next →

Example Question #11 : Solving Logarithmic Equations

Solve the equation.

$5^{11-3x}=125^{5-x}$

Possible Answers:

x=1

No solution

x=11

x=-11

x=15

Correct answer:

No solution

Explanation:

Change 125 to 5^3 so that both sides have the same base. Apply log and then set the exponential expressions equal to each other so that 11-3x=15-3x . Upon trying to isolate , it becomes clear that there is no solution.

Report an Error

Example Question #21 : Logarithms

Solve for :

$2 + \log_{3} (x - 2) = \log_{3} (x +6)$

Possible Answers:

$x = 6 \textrm{ or }x=10$

x = 3

The equation has no solution.

x = 10

x = 6

Correct answer:

x = 3

Explanation:

Since $3^{2} = 9$ , $2 = \log_{3}9$ , and we can rewrite and solve this statement as follows:

$2 + \log_{3} (x - 2) = \log_{3} (x +6)$

$\log_{3}9 + \log_{3} (x - 2) = \log_{3} (x +6)$

$\log_{3} 9 (x - 2) = \log_{3} (x +6)$

$3 ^ { \log_{3} 9 (x - 2)} =3 ^ { \log_{3} (x +6)}$

9(x-2) = x + 6

9x-18 = x + 6

9x-18 -x + 18 = x + 6 -x + 18

8x = 24

x = 3

Substitution confirms this to be the only solution.

Report an Error

Example Question #13 : Solving Logarithmic Equations

Solve for :

$\log_{4} (x-3) = \log_{16} (x-1)$

Possible Answers:

x = 5

$x = 2 \textrm{ or } x = 5$

The equation has no solution.

The solution set of the equation is the set of all real numbers.

x = 2

Correct answer:

x = 5

Explanation:

We can rewrite this as follows:

$\log_{4} (x-3) = \log_{16} (x-1)$

$\log_{4} (x-3) =\frac{ \log_{4} (x-1)}{ \log_{4} 16}$

$\log_{4} (x-3) =\frac{ \log_{4} (x-1)}{ 2}$

$2 \log_{4} (x-3) =\log_{4} (x-1)$

$\log_{4} \left [(x-3)^2\right ] =\log_{4} (x-1)$

$\dpi{120} 4^ {\log_{4} \left [(x-3)^2\right ]} =4^ { \log_{4} (x-1)}$

(x-3)^2 = x -1

x^2 - 6x + 9 = x -1

x^2 - 6x + 9-x +1 = x-1 -x +1

x^2 - 7x + 10 = 0

(x-2)(x-5) = 0

Split this into a compound statement, and solve:

$x - 2 \Rightarrow x = 2$

$x - 5 \Rightarrow x = 5$

Test both of these possible solutions by substituting:

x = 2:

$\log_{4} (x-3) = \log_{16} (x-1)$

$\log_{4} (2-3) = \log_{16} (2-1)$

$\log_{4} (-1) = \log_{16} 1$

Since only positive numbers can have logarithms, this equation does not make sense. x = 2 is not a solution.

x = 5:

$\log_{4} (x-3) = \log_{16} (x-1)$

$\log_{4} (5-3) = \log_{16} (5-1)$

$\log_{4} 2 = \log_{16} 4$

$\frac{1}{2} =\frac{1}{2}$

x = 5 is a solution - the only solution.

Report an Error

Example Question #14 : Solving Logarithmic Equations

What is the sum of the value(s) of x that satisfy the equation $\log_{10}(2x^2-35x)=2$ ?

Possible Answers:

100

35/2

No solution.

Correct answer:

35/2

Explanation:

$\log_{10}(2x^2-35x)=2$

We need to rewrite the logarithmic equation into a form that is more recognizable.

An equation in the form $\log_{a}b=c$ can be rewritten in the exponential form b=a^c , where a, b, and c are constants (and b > 0). Thus, we will rewrite our original equation as follows:

2x^2-35x=10^2 .

We can now approach this as we would a typical quadratic equation.

2x^2-35x=10^2=100

Subtract 100 from both sides. We need to set all of the terms equal to zero.

2x^2-35x-100=0 .

We can factor this by first multiplying the outer coefficients. The product of 2 and -100 is equal to -200. We must think of two numbers that multiply to give -200 but add to give -35 (the middle coefficient). Two numbers that satisfy both of these requirements are -40 and 5. We will now rewrite the quadratic polynomial.

2x^2-35x-100=(2x^2-40x)+(5x-100)=0

We can use grouping to factor the polynomial. Factor the first two terms and the last two terms separately.

(2x^2-40x)+(5x-100)=0

2x(x-20)+5(x-20)=0

Notice that we can now factor out x - 20 from the 2x term and the 5 term.

(2x+5)(x-20)=0

To solve this, we set each factor equal to zero.

2x+5=0

$x=\frac{-5}{2}$

x-20=0

x=20.

The question ultimately asks for the sum of the values of x that satisfy the equation, so we must add -5/2 and 20, which yields 35/2.

The answer is 35/2.

Report an Error

Example Question #15 : Solving Logarithmic Equations

Solve the following logarithmic equation:

$log_{9}(x+4)+log_{9}(x-4)=1$

Possible Answers:

x=6

x=-5

x=5

x=1

x=4

Correct answer:

x=5

Explanation:

Since the bases of the logs are the same and the logarithms are added, the arguments can be multiplied together.

$log_{9}(x+4)+log_{9}(x-4)=1$

$log_{9}(x+4)(x-4)=1$

The logarithm can then be converted to exponential form.

$(x+4)(x-4)=9^{1}$

x^2-16=9

x^2-25=0

(x-5)(x+5)=0

x=5

Although there are two solutions to the equation, logarithims cannot be negative. Therefore, the only real solution is .

Report an Error

Example Question #16 : Solving Logarithmic Equations

Solve the following logarithmic equation:

$log_{2}[\left log_{9}(log_{3}x) \right ]=-1$

Possible Answers:

x=9

x=3

x=36

x=18

x=27

Correct answer:

x=27

Explanation:

Start by converting the logarithms into exponential form:

$log_{2}[\left log_{9}(log_{3}x) \right ]=-1$

$log_{9}(log_{3}x) =2^{-1}$

$log_{9}(log_{3}x) =\frac{1}{2}$

Convert again into exponential form:

$log_{3}x =9^{\frac{1}{2}}$

Simplify and solve:

$log_{3}x =(3^2)^{\frac{1}{2}}$

$log_{3}x =3$

x=3^3=27

Report an Error

Example Question #22 : Logarithms

Solve the following logarithmic equation:

$log_{2}(3x-2)=log_{2}(2x+6)$

Possible Answers:

x=10

x=5

x=7

x=8

x=6

Correct answer:

x=8

Explanation:

Since the bases of the logs are the same, the terms inside the logs can be set equal to each other.

3x-2=2x+6

x=8

Report an Error

Example Question #18 : Solving Logarithmic Equations

Solve the equation.

$2^{9x-3}=\left(\frac{1}{32}\right)^{x-11}$

Possible Answers:

$x=\frac{1}{9}$

$x=\frac{22}{14}$

$x=-\frac{26}{7}$

$x=\frac{22}{7}$

$x=\frac{29}{7}$

Correct answer:

$x=\frac{29}{7}$

Explanation:

Change the right side to $2^{-5}$ so that both sides are the same. Apply log to both sides so that you can set the exponential expressions equal to each other ( 9x-3=-5x+55 ).

$x=\frac{29}{7}$

Report an Error

← Previous 1 2 Next →

View Tutors

Richard
Certified Tutor

CUNY Brooklyn College, Bachelor of Science, Chemistry. CUNY Brooklyn College, Master of Arts, Experimental Psychology.

View Tutors

Carl
Certified Tutor

University of Georgia, Bachelors, English. Yale University, PHD, Medieval Studies.

View Tutors

Raphael
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Computer Science.

All High School Math Resources

8 Diagnostic Tests 613 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Calculus Tutors in San Francisco-Bay Area, SAT Tutors in Dallas Fort Worth, Spanish Tutors in Boston, Statistics Tutors in Philadelphia, Calculus Tutors in New York City, Statistics Tutors in Washington DC, LSAT Tutors in Philadelphia, Math Tutors in Boston, Biology Tutors in Washington DC, Math Tutors in San Francisco-Bay Area

Popular Courses & Classes

ISEE Courses & Classes in Chicago, ACT Courses & Classes in Chicago, Spanish Courses & Classes in Boston, MCAT Courses & Classes in Seattle, GMAT Courses & Classes in Los Angeles, GMAT Courses & Classes in San Diego, GMAT Courses & Classes in Miami, SAT Courses & Classes in Houston, MCAT Courses & Classes in Houston, MCAT Courses & Classes in Denver

Popular Test Prep

LSAT Test Prep in Washington DC, SSAT Test Prep in Seattle, ACT Test Prep in New York City, SAT Test Prep in Los Angeles, ACT Test Prep in Boston, GMAT Test Prep in Miami, SSAT Test Prep in New York City, LSAT Test Prep in Miami, MCAT Test Prep in Dallas Fort Worth, ACT Test Prep in Atlanta

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

High School Math : Solving and Graphing Logarithmic Equations

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #11 : Solving Logarithmic Equations

Example Question #21 : Logarithms

Example Question #13 : Solving Logarithmic Equations

Example Question #14 : Solving Logarithmic Equations

Example Question #15 : Solving Logarithmic Equations

Example Question #16 : Solving Logarithmic Equations

Example Question #22 : Logarithms

Example Question #18 : Solving Logarithmic Equations

All High School Math Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: