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Solving Logarithmic Equations: A Comprehensive Guide
Unlock the secrets of logarithmic equations with our step-by-step guide. Learn essential techniques for simplifying, solving, and mastering log problems to excel in advanced mathematics.

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Now Playing:Solve logarithmic equations– Example 0
Intros
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  1. How to Solve a Logarithmic Equation
Examples
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  1. Solve algebraically:
    1. log⁡(3−x)+log⁡(4−3x)−log⁡(x)=log⁡7 \log(3 - x) + \log (4 - 3x) - \log(x) = \log7

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    2. 2log⁡3(x+4)−log⁡3(−x)=2 2 \log_3 {(x + 4)} - \log_3 (- x) = 2

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    3. log⁡2x=2+12log⁡2(x−3)\log_2x = 2 + {1\over2} \log_2(x - 3)

Practice
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Solve Logarithmic Equations 1a
What is a logarithm?
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Notes

Solving Logarithmic Equations

Rules or Laws of Logarithms:

As you know, a logarithm is a mathematical operation that is the inverse of exponentiation. It is expressed by using the abbreviation "log". Before getting into solving logarithmic equations, there are several strategies and "rules" that we must first familiarize ourselves with.

First of all, in order to solve logarithmic equations, just like with polynomials, you should be comfortable graphing logarithmic functions. Check out our video on graphing logarithmic functions for an overview if needed. Also, before we get into logarithm rules, it is important that you also understand one of the simplest logarithm strategies – the change of base formula. Again, check out our video on the change of base formula if you need a refresher. Now that you have all that mastered, let's take a look at some of the most important logarithm rules:

1) Logarithm Product Rule

In general, the product rule of logarithms is defined by:

log⁡A+log⁡B=log⁡(A×B)\log A + \log B = \log (A \times B)

That is, when adding two logs of the same base, you can rewrite the expression as a single log by multiplying the terms within the logarithmic expression.

2) Logarithm Quotient Rule

In general, the quotient rule of logarithms is defined by:

log⁡A−log⁡B=log⁡(AB)\log A - \log B = \log (\frac{A}{B})

That is, when subtracting two logs of the same base, you can rewrite the expression as a single log by dividing the terms within the logarithmic expression.

3) Logarithm Power Rule

In general, the power rule of logarithms is defined by:

log⁡(A)B=B×log⁡A\log (A)^{B} = B \times \log A

That is, when there is an exponent on the term within the logarithmic expression, you can bring down that exponent and multiply it by the log.

4) Log of Exponent Rule

In general, the log of exponent rule is defined by:

log⁡A(AB)=B\log_{A} (A^{B}) = B

That is, when there is an exponent on the term within the logarithmic expression, and that term is the same as the base of the logarithm, the answer is simply the exponent.

5) Exponent of Log Rule

In general, the exponent of log rule is defined by:

Alog⁡A(B)=BA^{\log_{A} (B)} = B

That is, raising a logarithm of a number by its base equals that number.

6) Log Identity Rule

In general, the identity rule of logarithms is defined by:

log⁡AA=1\log_{A} A = 1

That is, when taking the log of something to the base of that same thing, the logarithmic expression is simply equal to just 1.

7) Special Logs

Though not necessarily rules, there are a couple of logs that you should know by heart to make things a little easier. They are:

log⁡1=0\log 1 = 0

log⁡0=undefined\log 0 = undefined

Both of these cases are always true, regardless of the base. Also, in case it comes up, the first special case is sometimes referred to as the logarithmic zero rule.

All of these rules, taken together, are extremely powerful tools we can use to solve any logarithmic problem. For a video review of these concepts, check out our videos on properties of logarithms and the quotient rule for logarithms. Now that we've covered the essentials, let's get to how to solve log problems!

How to Solve Log Problems:

As with anything in mathematics, the best way to learn how to solve log problems is to do some practice problems! We will use the rules we have just discussed to solve some examples.

Example 1:

Solve the logarithmic equation:

log⁡(3−x)+log⁡(4−3x)−log⁡(x)=log⁡7\log (3 - x) + \log (4 - 3x) - \log (x) = \log 7

Step 1: Use Known Log Rules

In any problem that involves solving logarithmic equations, the first step is to always try to simplify using the log rules. In this case, we will use the product, quotient, and exponent of log rules. We do this to try to make a polynomial/algebraic equation that is easier to solve. This is shown below:

log⁡[(3−x)(4−3x)x]=log⁡7\log [\frac{(3-x)(4-3x)}{x}] = \log 7

10log⁡[(3−x)(4−3x)x]=10log⁡710^{\log [\frac{(3-x)(4-3x)}{x}]} = 10^{\log 7}

(3−x)(4−3x)x=7\frac{(3-x)(4-3x)}{x} = 7

Step 2: Solve Equation

We are left with an algebraic equation which we can now solve.

3x2−13x+12x=7\frac{3x^{2} - 13x + 12}{x} = 7

3x2−13x+12=7x3x^{2} - 13x + 12 = 7x

3x2−20x+12=03x^{2} - 20x + 12 = 0

(3x−2)(x−6)=0(3x - 2)(x - 6) = 0

3x−2=0orx−6=03x - 2 = 0 or x - 6 = 0

x=23orx=6x = \frac{2}{3} or x = 6

Step 3: Check Solutions

Because we initially had a logarithmic equation, we need to check our answers to make sure they are valid.

log⁡(3−23)+log⁡(4−3(23))−log⁡(23)=log⁡7\log (3 - \frac{2}{3}) + \log (4 - 3(\frac{2}{3})) - \log (\frac{2}{3}) = \log 7

The solution x=23x = \frac{2}{3} is correct.

log⁡(3−6)+log⁡(4−3(6))−log⁡(6)=log⁡7\log (3 - 6) + \log (4 - 3(6)) - \log (6) = \log 7

The solution x=6x = 6 is rejected because the log of a negative number is undefined.

Example 2:

Step 1: Use Known Log Rules

In this case, we will use the power and quotient log rules. We do this to try to make a polynomial/algebraic equation that is easier to solve. This is shown below:

2log⁡3(x+4)−log⁡3(−x)=22\log_{3} (x + 4) - \log_{3} (-x) = 2

log⁡3(x+4)2−log⁡3(−x)=2\log_{3} (x + 4)^{2} - \log_{3} (-x) = 2

log⁡3(x2+8x+16−x)=2\log_{3} (\frac{x^{2} + 8x + 16}{-x}) = 2

Step 2: Simplify

We can convert to exponent form because one side has log and the other side does not.

32=x2+8x+16−x3^{2} = \frac{x^{2} + 8x + 16}{-x}

Step 3: Solve Equation

We are left with an algebraic equation which we can now solve.

−9x=x2+8x+16-9x = x^{2} + 8x + 16

x2+17x+16=0x^{2} + 17x + 16 = 0

(x+16)(x+1)=0(x + 16)(x + 1) = 0

x=−16orx=−1x = -16 or x = -1

Step 4: Check Solutions

Because we initially had a logarithmic equation, we need to check our answers to make sure they are valid.

2log⁡3((−16)+4)−log⁡3(−(−16))=22\log_{3} ((-16) + 4) - \log_{3} (-(-16)) = 2

The solution x = -16 is rejected.

2log⁡3((−1)+4)−log⁡3(−(−1))=22\log_{3} ((-1) + 4) - \log_{3} (-(-1)) = 2

The solution x = -1 is correct.

Example 3:

Step 1: Simplify

Multiply both sides of the equation by 2 to get rid of the fraction.

log⁡2x=2+12log⁡2(x−3)\log_{2} x = 2 + \frac{1}{2} \log_{2} (x - 3)

2log⁡2x=4+log⁡2(x−3)2\log_{2} x = 4 + \log_{2} (x - 3)

Step 2: Use Known Log Rules

In this case, we will use the power of log and quotient log rules. We can then simplify like in the previous example to make the exponential form. We do this to try to make a polynomial/algebraic equation that is easier to solve. This is shown below:

log⁡2(x)2=4+log⁡2(x−3)\log_{2} (x)^{2} = 4 + \log_{2} (x - 3)

log⁡2(x)2−log⁡2(x−3)=4\log_{2} (x)^{2} - \log_{2} (x - 3)= 4

log⁡2(x2x−3)=4\log_{2} (\frac{x^{2}}{x-3}) = 4

24=x2x−32^{4} = \frac{x^{2}}{x-3}

Step 3: Solve Equation

We are left with an algebraic equation which we can now solve.

16=x2x−316 = \frac{x^{2}}{x-3}

16x−48=x216x - 48 = x^{2}

x2−16x+48=0x^{2} - 16x + 48 = 0

(x−4)(x−12)=0(x - 4)(x - 12) = 0

x=4orx=12x = 4 or x = 12

Step 4: Check Solutions

Because we initially had a logarithmic equation, we need to check our answers to make sure they are valid.

log⁡24=2+12log⁡2(4−3)\log_{2} 4 = 2 + \frac{1}{2} \log_{2} (4 - 3)

The solution x = 4 checks out.

log⁡212=2+12log⁡2(12−3)\log_{2} 12 = 2 + \frac{1}{2} \log_{2} (12 - 3)

So does x=12. In this problem, we get to keep both our answers.

Example 4:

Step 1: Use Known Log Rules

In this case, we will use the exponent of log rule. We do this to try to make a polynomial/algebraic equation that is easier to solve. Note: (log⁡x)2(\log x)^{2} is different than log⁡x2\log x^{2}, and thus we cannot simplify the first log⁡\log.This is shown below:

(log⁡x)2−log⁡x5=14(\log x)^{2} - \log x^{5} = 14

(log⁡x)2−5log⁡x=14(\log x)^{2} - 5\log x = 14

Step 2: Substitution

To make this equation easier to solve, we can substitute log⁡x\log x as "a" to make a quadratic equation!

a2−5a=14a^{2} - 5a = 14

a2−5a−14=0a^{2} - 5a - 14 = 0

(a−7)(a+2)=0(a - 7)(a + 2) = 0

a=7ora=−2a = 7 or a = -2

*** Since we used substitution, we need to replace "a" back with the original term! ***

log⁡x=7orlog⁡x=−2\log x = 7 or \log x = -2

107=xor10−2=x10^{7} = x or 10^{-2} = x

Step 4: Check Solutions

Because we initially had a logarithmic equation, we need to check our answers to make sure they are valid.

(log⁡107)2−log⁡(107)5=14(\log 10^{7})^{2} - \log (10^{7})^{5} = 14

The solution x=107x = 10^{7} is correct.

(log⁡10−2)2−log⁡(10−2)5=14(\log 10^{-2})^{2} - \log (10^{-2})^{5} = 14

The solution x=10−2x = 10^{-2} is not correct.

And that's all there is too it! To check your work with future practice problems, be sure to use this excellent calculator here. Lastly, for a video review of everything we've just covered, check out our video on how to solve log equations.

Steps to solving logarithmic equations
1. Combine all the logarithms into one.
2. Two scenarios:
i.
log⁡bM=log⁡bN\log_bM = \log_bN →M=N M=N
ii. log⁡bM=N\log_bM = N →bN=M b^N = M
3. Plug the answer back into the original equation to make sure the inside of any logarithm is non-negative.