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## Solving Logarithmic Equations

## Solving logarithmic equations

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- How to Solve a Logarithmic Equation
- log ( 3 − x ) + log ( 4 − 3 x ) − log ( x ) = log 7 \log(3 - x) + \log (4 - 3x) - \log(x) = \log7 lo g ( 3 − x ) + lo g ( 4 − 3 x ) − lo g ( x ) = lo g 7
- 2 log 3 ( x + 4 ) − log 3 ( − x ) = 2 2 \log_3 {(x + 4)} - \log_3 (- x) = 2 2 lo g 3 ( x + 4 ) − lo g 3 ( − x ) = 2
- log 2 x = 2 + 1 2 log 2 ( x − 3 ) \log_2x = 2 + {1\over2} \log_2(x - 3) lo g 2 x = 2 + 2 1 lo g 2 ( x − 3 )
- ( log x ) 2 − log x 5 = 14 {(\log x)^2 - \log x^5 = 14} ( lo g x ) 2 − lo g x 5 = 14
- 2 ( log 3 n ) 3 − ( log 3 n ) 2 = 0 {2 (\log_3n)^3 - (\log_3n)^2 = 0} 2 ( lo g 3 n ) 3 − ( lo g 3 n ) 2 = 0

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## Topic Notes

Solving logarithmic equations, rules or laws of logarithms:.

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

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

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

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

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

log A ( A B ) = B \log_{A} (A^{B}) = B lo g A ( A B ) = B

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

A log A ( B ) = B A^{\log_{A} (B)} = B A l o g A ( B ) = B

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

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

log A A = 1 \log_{A} A = 1 lo g A A = 1

log 1 = 0 \log 1 = 0 lo g 1 = 0

log 0 = u n d e f i n e d \log 0 = undefined lo g 0 = u n d e f in e d

## How to Solve Log Problems:

Solve the logarithmic equation:

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

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

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

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

3 x 2 − 20 x + 12 = 0 3x^{2} - 20x + 12 = 0 3 x 2 − 20 x + 12 = 0

( 3 x − 2 ) ( x − 6 ) = 0 (3x - 2)(x - 6) = 0 ( 3 x − 2 ) ( x − 6 ) = 0

3 x − 2 = 0 o r x − 6 = 0 3x - 2 = 0 or x - 6 = 0 3 x − 2 = 0 or x − 6 = 0

x = 2 3 o r x = 6 x = \frac{2}{3} or x = 6 x = 3 2 or x = 6

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

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

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

3 2 = x 2 + 8 x + 16 − x 3^{2} = \frac{x^{2} + 8x + 16}{-x} 3 2 = − x x 2 + 8 x + 16

− 9 x = x 2 + 8 x + 16 -9x = x^{2} + 8x + 16 − 9 x = x 2 + 8 x + 16

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

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

x = − 16 o r x = − 1 x = -16 or x = -1 x = − 16 or x = − 1

The solution x = -16 is rejected.

The solution x = -1 is correct.

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

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

2 4 = x 2 x − 3 2^{4} = \frac{x^{2}}{x-3} 2 4 = x − 3 x 2

16 = x 2 x − 3 16 = \frac{x^{2}}{x-3} 16 = x − 3 x 2

16 x − 48 = x 2 16x - 48 = x^{2} 16 x − 48 = x 2

x 2 − 16 x + 48 = 0 x^{2} - 16x + 48 = 0 x 2 − 16 x + 48 = 0

( x − 4 ) ( x − 12 ) = 0 (x - 4)(x - 12) = 0 ( x − 4 ) ( x − 12 ) = 0

x = 4 o r x = 12 x = 4 or x = 12 x = 4 or x = 12

The solution x = 4 checks out.

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

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

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

a 2 − 5 a = 14 a^{2} - 5a = 14 a 2 − 5 a = 14

a 2 − 5 a − 14 = 0 a^{2} - 5a - 14 = 0 a 2 − 5 a − 14 = 0

( a − 7 ) ( a + 2 ) = 0 (a - 7)(a + 2) = 0 ( a − 7 ) ( a + 2 ) = 0

a = 7 o r a = − 2 a = 7 or a = -2 a = 7 or a = − 2

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

log x = 7 o r log x = − 2 \log x = 7 or \log x = -2 lo g x = 7 or lo g x = − 2

1 0 7 = x o r 1 0 − 2 = x 10^{7} = x or 10^{-2} = x 1 0 7 = x or 1 0 − 2 = x

The solution x = 1 0 7 x = 10^{7} x = 1 0 7 is correct.

The solution x = 1 0 − 2 x = 10^{-2} x = 1 0 − 2 is not correct.

## Basic Concepts

- Evaluating logarithms using change-of-base formula
- Product rule of logarithms
- Quotient rule of logarithms

## Related Concepts

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## Solving Logarithmic Equations

## Types of Logarithmic Equations

Let’s learn how to solve logarithmic equations by going over some examples.

## Examples of How to Solve Logarithmic Equations

Example 1: Solve the logarithmic equation.

- Apply Product Rule from Log Rules .
- Distribute: \left( {x + 2} \right)\left( 3 \right) = 3x + 6
- Drop the logs, set the arguments (stuff inside the parenthesis) equal to each other.
- Then solve the linear equation. I know you got this part down!

Just a big caution. ALWAYS check your solved values with the original logarithmic equation.

- It is OKAY for x to be 0 or negative.
- However, it is NOT ALLOWED to have a logarithm of a negative number or a logarithm of zero, 0 , when substituted or evaluated into the original logarithm equation.

CAUTION: The logarithm of a negative number, and the logarithm of zero are both not defined .

{\log _b}\left( {{\rm{negative\,\,number}}} \right) = {\rm{undefined}}

{\log _b}\left( 0 \right) = {\rm{undefined}}

Yes! Since x = 7 checks, we have a solution at \color{blue}x = 7 .

Example 2: Solve the logarithmic equation.

- Apply Product Rule from Log Rules
- Simplify: \left( x \right)\left( {x - 2} \right) = {x^2} - 2x
- Drop the logs, set the arguments (stuff inside the parenthesis) equal to each other
- Solve the quadratic equation using the factoring method . But you need to move everything on one side while forcing the opposite side equal to 0 .
- Set each factor equal to zero, then solve for x .

x + 2 = 0 implies that x = - 2

Let’s check our potential answers x = 5 and x = - 2 if they will be valid solutions.

Example 3: Solve the logarithmic equation.

- The difference of logs is telling us to use the Quotient Rule . Convert the subtraction operation outside into a division operation inside the parenthesis. Do it to both sides of the equations.
- I think we are ready to set each argument equal to each other since we can reduce the problem to have a single log expression on each side of the equation.
- Drop the logs, and set the arguments (stuff inside the parenthesis) equal to each other. Note that this is a Rational Equation . One way to solve it is to get its Cross Product .
- It looks like this after getting its Cross Product.
- Simplify both sides by the Distributive Property. At this point, we realize that it is just a Quadratic Equation. No big deal then. Move everything to one side, which forces one side of the equation to be equal to zero.
- This is easily factorable. Now set each factor to zero and solve for x .
- So, these are our possible answers.

Example 4: Solve the logarithmic equation.

Well, we have to bring it up as an exponent using the Power Rule in reverse.

- Bring up that coefficient \large{1 \over 2} as an exponent (refer to the leftmost term)
- Simplify the exponent (still referring to the leftmost term)
- Then, condense the logs on both sides of the equation. Use the Quotient Rule on the left and Product Rule on the right.
- Here, I used different colors to show that since we have the same base (if not explicitly shown it is assumed to be base 10 ), it’s okay to set them equal to each other.
- Dropping the logs and just equating the arguments inside the parenthesis.
- At this point, you may solve the Rational Equation by performing Cross Product. Move all the terms on one side of the equation, then factor them out.
- Set each factor equal to zero and solve for x .

Example 5: Solve the logarithmic equation.

This problem involves the use of the symbol \ln instead of \log to mean logarithm.

Think of \ln as a special kind of logarithm using base e where e \approx 2.71828 .

- Use Product Rule on the right side
- Write the variable first, then the constant to be ready for the FOIL method .
- Simplify the two binomials by multiplying them together.
- At this point, I simply color-coded the expression inside the parenthesis to imply that we are ready to set them equal to each other.
- Yep! This is where we say that the stuff inside the left parenthesis equals the stuff inside the right parenthesis.
- Solve the Quadratic Equation using the Square Root Method . You do it by isolating the squared variable on one side and the constant on the other. Then we apply the square root on both sides.

Example 6: Solve the logarithmic equation.

We will transform the equation from the logarithmic form to the exponential form, then solve it.

- I color-coded the parts of the logarithmic equation to show where they go when converted into exponential form.
- The blue expression stays at its current location, but the red number becomes the exponent of the base of the logarithm which is 3 .
- Simplify the right side, {3^4} = 81 .
- Finish off by solving the two-step linear equation that arises.

Example 7: Solve the logarithmic equation.

- Move all the logarithmic expressions to the left of the equation, and the constant to the right.
- Use the Quotient Rule to condense the log expressions on the left side.
- Get ready to write the logarithmic equation into its exponential form.
- The blue expression stays in its current location, but the red constant turns out to be the exponent of the base of the log.
- Simplify the right side of the equation since 5^{\color{red}1}=5 .
- This is a Rational Equation due to the presence of variables in the numerator and denominator.

Example 8: Solve the logarithmic equation.

- Move the log expressions to the left side, and keep the constant to the right.
- Apply the Quotient Rule since they are the difference of logs.
- I used different colors here to show where they go after rewriting in exponential form.
- Notice that the expression inside the parenthesis stays in its current location, while the \color{red}5 becomes the exponent of the base.
- To solve this Rational Equation, apply the Cross Product Rule.
- Simplify the right side by the distributive property . It looks like we are dealing with a quadratic equation.
- Move everything to the left side and make the right side just zero.

Factor out the trinomial. Set each factor equal to zero then solve for x .

Make sure that you check the potential answers from the original logarithmic equation.

Example 9: Solve the logarithmic equation

- Let’s keep the log expressions on the left side while the constant on the right side.
- Start by condensing the log expressions using the Product Rule to deal with the sum of logs.
- Then further condense the log expressions using the Quotient Rule to deal with the difference of logs.
- At this point, I used different colors to illustrate that I’m ready to express the log equation into its exponential equation form.
- Keep the expression inside the grouping symbol ( blue ) in the same location while making the constant \color{red}1 on the right side as the exponent of the base 7 .
- Solve this Rational Equation using Cross Product. Express 7 as \large{7 \over 1} .
- Cross multiply.
- Move all terms to the left side of the equation. Factor out the trinomial. Next, set each factor equal to zero and solve for x .
- These are your potential answers. Always check your values.

Thus, the only solution is \color{blue}x=11 .

Example 10: Solve the logarithmic equation.

- Keep the log expression on the left, and move all the constants on the right side.
- I think we’re ready to transform this log equation into the exponential equation.
- The expression inside the parenthesis stays in its current location while the constant 3 becomes the exponent of the log base 3 .
- Simplify the right side since {3^3}=27 . What we have here is a simple Radical Equation .

- To get rid of the radical symbol on the left side, square both sides of the equation.
- After squaring both sides, it looks like we have a linear equation. Just solve it as usual.

Check your potential answer back into the original equation.

After doing so, you should be convinced that indeed \color{blue}x=-104 is a valid solution.

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## How to Solve Logarithms

Last Updated: March 29, 2019 References

## Before You Begin: Learn to Express a Logarithmic Equation Exponentially [1] X Research source [2] X Research source

- If and only if: b y = x
- b does not equal 1
- In the same equation, y is the exponent and x is the exponential expression that the logarithm is set equal to.

## Method One: Solve for X

- Comparing this equation to the definition [ y = log b (x) ], you can conclude that: y = 4; b = 3; x = x + 5
- Rewrite the equation so that: b y = x
- 3 4 = x + 5

## Method Two: Solve for X Using the Logarithmic Product Rule [3] X Research source [4] X Research source

- Comparing this equation to the definition [ y = log b (x) ], you can conclude that: y = 2; b = 4 ; x = x 2 + 6x
- 4 2 = x 2 + 6x

- 4 * 4 = x 2 + 6x
- 16 = x 2 + 6x
- 16 - 16 = x 2 + 6x - 16
- 0 = x 2 + 6x - 16
- 0 = (x - 2) * (x + 8)
- x = 2; x = -8

- Example: x = 2
- Note that you cannot have a negative solution for a logarithm, so you can discard x - 8 as a solution.

## Method Three: Solve for X Using the Logarithmic Quotient Rule [5] X Research source

- Comparing this equation to the definition [ y = log b (x) ], you can conclude that: y = 2; b = 3; x = (x + 6) / (x - 2)
- 3 2 = (x + 6) / (x - 2)

- 3 * 3 = (x + 6) / (x - 2)
- 9 = (x + 6) / (x - 2)
- 9 * (x - 2) = [(x + 6) / (x - 2)] * (x - 2)
- 9x - 18 = x + 6
- 9x - x - 18 + 18 = x - x + 6 + 18
- 8x / 8 = 24 / 8

## Community Q&A

## You Might Also Like

- ↑ http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut43_logfun.htm#logdef
- ↑ http://www.mathsisfun.com/algebra/logarithms.html
- ↑ http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut46_logeq.htm
- ↑ http://dl.uncw.edu/digilib/mathematics/algebra/mat111hb/eandl/equations/equations.html
- ↑ http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut44_logprop.htm

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## How To Solve Logarithms in 8 Steps (With Examples)

## What is a logarithm?

## Uses of logarithms

Consider these important uses of logarithms across different applications:

## Supports statistical research

## Simplifies complex exponential problems

## Gives insight into growth and decay rates

## How to solve logs in expressions

Use the following steps to simplify logarithmic expressions:

## 1. Identify the base and the power

## 2. Simplify by multiplying

## 3. Apply the process to larger expressions

## 4. Use the variable rules

## How to solve logs in equations

Use the steps below to solve logs in equations:

## 1. Apply the log rules to combine like terms

## 2. Simplify the resulting mathematical operations

## 3. Balance the equation to isolate the variable

## 4. Use the result to check the log

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## Properties of logarithmic functions

For example , 32 = 2 × 2 × 2 × 2 × 2 = 2 2 .

Then the logarithmic function is given by;

f(x) = log b x = y, where b is the base, y is the exponent, and x is the argument.

## How to Solve Logarithmic Functions?

To solve an equation with logarithm(s), it is important to know their properties.

Some of the properties are listed below.

⟹ log a (p q) = log a p + log a q.

⟹ log a (p/q) = log a p – log a q

Other properties of logarithmic functions include:

- The bases of an exponential function and its equivalent logarithmic function are equal.
- The logarithms of a positive number to the base of the same number are equal to 1.

- Log a 0 is undefined
- Logarithms of negative numbers are undefined.
- The base of logarithms can never be negative or 1.
- A logarithmic function with base 10is called a common logarithm. Always assume a base of 10 when solving with logarithmic functions without a small subscript for the base.

Let’s use these properties to solve a couple of problems involving logarithmic functions.

Rewrite exponential function 7 2 = 49 to its equivalent logarithmic function.

Write the logarithmic equivalent of 5 3 = 125.

If 2 log x = 4 log 3, then find the value of ‘x’.

Find the logarithm of 1024 to the base 2.

Find the value of x in log 2 ( x ) = 4

Rewrite the logarithmic function log 2 ( x ) = 4 to exponential form.

Solve for x in the following logarithmic function log 2 (x – 1) = 5.

Solution Rewrite the logarithm in exponential form as;

log 2 (x – 1) = 5 ⟹ x – 1 = 2 5

Now, solve for x in the algebraic equation. ⟹ x – 1 = 32 x = 33

Find the value of x in log x 900 = 2.

Write the logarithm in exponential form as;

Find the square root of both sides of the equation to get;

But since, the base of logarithms can never be negative or 1, therefore, the correct answer is 30.

Solve for x given, log x = log 2 + log 5

Using the product rule Log b (m n) = log b m + log b n we get;

⟹ log 2 + log 5 = log (2 * 5) = Log (10).

Rewrite the logarithm in exponential form to get;

Now, solve the quadratic equation. x 2 = 4x – 3 x 2 – 4x + 3 = 0 (x -1) (x – 3) = 0

Since the base of a logarithm can never be 1, then the only solution is 3.

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## Solving Log Equations with Exponentials

Using the Definition Using Exponentials Calculators & Etc.

The second type of log equation requires the use of The Relationship :

...........is equivalent to............ (means the exact same thing as)

In animated form, the two equations are related as shown below:

## MathHelp.com

## Solve log 2 ( x ) = 4

## Solve log 2 (8) = x .

But 8 = 2 3 , so I can equate powers of two:

Note that this could also have been solved by working directly from the definition of a logarithm.

What power, when put on " 2 ", would give you an 8 ? The power 3 , of course!

## Solve log 2 ( x ) + log 2 ( x − 2) = 3

log 2 ( x ) + log 2 ( x − 2) = 3

## Solve log 2 (log 2 ( x )) = 1

The Relationship converts the above to:

Now I'll apply The Relationship a second time:

## Solve log 2 ( x 2 ) = (log 2 ( x )) 2

First, I'll expand the square on the right-hand side to be the explicit product of two logs:

log 2 ( x 2 ) = [log 2 ( x )] 2

log 2 ( x 2 ) = [log 2 ( x )] [log 2 ( x )]

2·log 2 ( x ) = [log 2 ( x )] [log 2 ( x )]

Then I'll move that term from the left-hand side of the equation to the right-hand side:

0 = [log 2 ( x )] [log 2 ( x )] − 2·log 2 ( x )

0 = [log 2 ( x )] [log 2 ( x ) − 2]

log 2 ( x ) = 0 or log 2 ( x ) − 2 = 0

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## Section 6.4 : Solving Logarithm Equations

Let’s take a look at a couple of examples.

- \(2{\log _9}\left( {\sqrt x } \right) - {\log _9}\left( {6x - 1} \right) = 0\)
- \(\log x + \log \left( {x - 1} \right) = \log \left( {3x + 12} \right)\)
- \(\ln 10 - \ln \left( {7 - x} \right) = \ln x\)

Now, before we declare these to be solutions we MUST check them in the original equation.

No logarithms of negative numbers and no logarithms of zero so this is a solution.

So, with all that out of the way, we’ve got a single solution to this equation, \(x = 6\).

We’ve got two possible solutions to check here.

- \({\log _5}\left( {2x + 4} \right) = 2\)
- \(\log x = 1 - \log \left( {x - 3} \right)\)
- \({\log _2}\left( {{x^2} - 6x} \right) = 3 + {\log _2}\left( {1 - x} \right)\)

So, let’s do that with this equation. The exponential form of this equation is,

Notice that this is an equation that we can easily solve.

Only positive numbers in the logarithm and so \(x = \frac{{21}}{2}\) is in fact a solution.

Here is the exponential form of this equation.

So, we’ve got two potential solutions. Let’s check them both.

We’ve got negative numbers in the logarithms and so this can’t be a solution.

No negative numbers or zeroes in the logarithms and so this is a solution.

Therefore, we have a single solution to this equation, \(x = 5\).

Again, let’s get the logarithms onto one side and combined into a single logarithm.

Now, convert it to exponential form.

Now, let’s solve this equation.

Now, let’s check both of these solutions in the original equation.

Therefore, we get a single solution for this equation, \(x = - 4\).

## IMAGES

## VIDEO

## COMMENTS

This algebra video tutorial explains how to solve logarithmic equations ... This video include examples and practice problems with natural

The steps for solving them follow. Step 1: Use the properties of the logarithm to isolate the log on one side. Step 2: Apply the definition of the logarithm and

How to Solve Log Problems: · 1. Combine all the logarithms into one. · 2. Two scenarios: i. log b M = log b N \log_bM = \log_bN logbM=logbN→ M = N M=N M=N.

Examples of How to Solve Logarithmic Equations · log base b of M times N is equal to log base b of M plus · log base 2 of x plus 2 plus log base 2 of 3 is equal

Method One: Solve for X ... Isolate the logarithm. Use inverse operations to move any part of the equation that is not part of the logarithm to the opposite side

How to solve logs in equations · 1. Apply the log rules to combine like terms · 2. Simplify the resulting mathematical operations · 3. Balance the

How to solve equations with logarithms on one side? · Simplify the logarithmic equations by applying the appropriate laws of logarithms. · Rewrite the logarithmic

The product rule of logarithm states the logarithm of the product of two numbers having a common base is equal to the sum of individual logarithms. ⟹ log a (

Solve log2(x) + log2(x − 2) = 3 · log2(x) + log2(x − 2) = 3. log2[(x)(x − 2)] = 3. log2(x2 − 2x) = 3. Now the equation is arranged in a useful way. · log2(x2

Section 6.4 : Solving Logarithm Equations · log5(2x+4)=2 log 5 ( 2 x + 4 ) = 2 · logx=1−log(x−3) log x = 1 − log ( x − 3 ) · log2(x2−6x)=3