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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, \(x = - 4\).

## Logarithmic Equations: Very Difficult Problems with Solutions

## Solving More Complex Logarithmic Equations

## More Topics

## Educational Videos

## 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.

You might also be interested in:

## OpenAlgebra.com

## Solving Logarithmic Equations

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## Solving Logarithmic Equations

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\( \log_x(a) + \log_x(b) = \log_x(ab) \)

\(\log_x(a) - \log_x(b) = \log_x\big(\frac ab\big) \)

\(\log_x(a) = \frac{\log_y(a)}{\log_y(x)} = \frac1{\log_a(x)}\) for any positive real number \(y \)

Common mistakes to watch out for include

\(\log_x(a) \cdot \log_x(b) \neq \log_x(ab) \)

\(\frac{\log_x(a)}{\log_x(b)} \neq \log_x\big(\frac{a}{b}\big)\).

## Solving Logarithmic Equations - Basic

Solving logarithmic equations - intermediate.

Find \( x \) if \( \log_2(3x+1) = 4 \). By the definition of the logarithm, \[\begin{align} 3x+1 &= 2^4 \\&= 16 \\ 3x &= 15 \\ x &= 5.\ _\square \end{align}\]

If \(\log_{8}m + \log_{8}\frac{1}{6}=\frac{2}{3}\), then what is \(m?\)

Find all \( x\) such that \( \log_3(x-12) +\log_3(x-6) = 3. \) Simplify the given equation as follows: \[\begin{align} \log_3(x-12) + \log_3(x-6) &=3 \\ \log_3\big((x-12)(x-6)\big) &= 3\\ (x-12)(x-6) &= 27 \\ x^2-18x-45 &=0\\ (x-15)(x-3) &= 0. \end{align}\] This produces two potential solutions \( x=3, x=15\). But note that \( x = 3 \) is not an actual solution, as \( \log_3(3-12) \) is undefined. The only actual solution is \( x=15 \). \(_\square\)

Find all real solutions \(x\) to

\[3\log_2(x) - 1 = \log_2\left(\frac32 x-1\right).\]

Enter your answer as the sum of all such \( x \).

Suppose \( a,b\) are positive real numbers such that \[\log_a(10)+\log_b(100) = \log_{ab}(1000000).\] Show that \( a = b \) or \( a^2=b \). Rewrite this as \[\frac{\log(10)}{\log(a)} + \frac{\log(100)}{\log(b)} = \frac{\log(1000000)}{\log(ab)},\] where the logs are all to the base \( 10 \). This simplifies to \[\frac1{\log(a)}+\frac2{\log(b)} = \frac6{\log(ab)}.\] Let \( m = \log(a) \) and \( n = \log(b) \). Then \(\log(ab) = m+n\), so \[\begin{align} \frac1{m}+\frac2{n} &= \frac6{m+n} \\ n+2m &= \frac{6mn}{m+n} \\ (2m+n)(m+n) &= 6mn \\ 2m^2+3mn+n^2 &= 6mn \\ 2m^2-3mn+n^2 &= 0 \\ (2m-n)(m-n) &= 0. \end{align}\] So \( m=n\) or \( 2m=n\). In the first case, \( \log(a)=\log(b) \), so \( a=b\). In the second case, \( 2\log(a) = \log(b) \), so \( \log(a^2)=\log(b) \), so \(a^2=b \). \(_\square \)

The sum of all (positive) solutions of the equation

\[\log_{16}x +\log _x 16=\log_{512} x + \log_x {512}\]

Equations involving exponents can often be simplified by taking logarithms:

Solve for \( x \) if \[\left( \frac{x}2 \right)^{\log_2(x)} = 8x.\] Take \(\log_2\) of both sides: \[\begin{align} \log_2\left(\left(\frac{x}2 \right)^{\log_2(x)}\right) &= \log_2(8x) \\ \log_2(x)\log_2\left(\frac x2\right) &= \log_2(x)+3 \\ \log_2(x)\big(\log_2(x)-1\big) &= \log_2(x)+3, \end{align}\] and substituting \( y = \log_2(x) \) turns this into \( y(y-1)=y+3\), or \( y^2-2y-3 = 0 \). So \( y = 3 \) or \( y = -1 \), which leads to \( x = 8 \) or \( x = \frac 12 \). \(_\square\)

Given that \[\begin{align} \log_{10}(2) &= 0.3010299\ldots \\ \log_{10}(3) &= 0.4771212\ldots \\ \log_{10}(7) &= 0.8450980\ldots, \end{align}\] which is bigger, \( 12^{50} \) or \( 7^{64} \)? How many decimal digits do these two numbers have? Compute \[\begin{align} \log_{10}\big(2^{100}3^{50}\big) &= 100\log_{10}(2)+50\log_{10}(3) \\ &= 30.10299\ldots + 23.85656\ldots \\ &= 53.9595\ldots \\\\ \log_{10}\big(7^{64}\big) &= 64\log_{10}(7) \\ &= 64(0.8450980\ldots) \\ &= 54.0862\ldots. \end{align}\] So \( 7^{64} \) is bigger. The number of decimal digits of \( x \) is \( \lfloor \log_{10}(x) \rfloor +1 \), so \( 2^{100}3^{50} \) has \( 54 \) digits and \( 7^{64} \) has \( 55 \) digits. \(_\square\)

Find the number of digits in \(\Large 9^{9^{2}}\).

Note : You may use the fact that \(\log_{10} 3 = 0.4771\) correct up to 4 decimal places.

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## Solving challenging logarithm equation

I came across this logarithm equation and can't seem to figure out how to solve it.

$$lg(2x-24)=2+ \frac 13 lg8 - \frac 14 x lg16$$

I only managed to simplify all the way till $$100=(x-12)(2^x)$$ (based on assumption of base 10)

Would gladly appreciate if anyone could suggest a trick to solve this.

- $\begingroup$ Just to confirm, is this supposed to be log base 10? $\endgroup$ – 2012ssohn Aug 17, 2017 at 5:10
- $\begingroup$ @2012ssohn I presume $\lg$ is base two logarithm, but then I cannot see where $100$ comes from. $\endgroup$ – Angina Seng Aug 17, 2017 at 5:15
- $\begingroup$ There is no obvious integer solution to your last equation, so I'm tempted to say it's unsolvable (equations usually are when you have $x$ both as an exponent and as a "non-exponent", for lack of a better word). $\endgroup$ – Arthur Aug 17, 2017 at 5:16
- $\begingroup$ @2012ssohn The question didn't state but I would believe it to be base 10. I used base 10 to obtain 100. $\endgroup$ – javabeginner_0101 Aug 17, 2017 at 5:17
- $\begingroup$ @Arthur What about if any real solutions can be accepted? $\endgroup$ – javabeginner_0101 Aug 17, 2017 at 5:20

## 3 Answers 3

- 1 $\begingroup$ Is it possible to solve without such numerical methods? Assuming knowledge of math up to high school level. $\endgroup$ – javabeginner_0101 Aug 17, 2017 at 5:27
- 1 $\begingroup$ Mixes of exponentials and polynomials are usually not soluable without the W function or numerics. $\endgroup$ – Ross Millikan Aug 17, 2017 at 5:30

This is indeed a very good introduction to Lambert function.

Once you arrive at $100$ = ( $x−12$ )( $2^x$ ) ,

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## IMAGES

## VIDEO

## COMMENTS

This video explains how to solve complex logarithmic equations using properties of logarithms such as the change of base formula, the power

Learn how to solve advanced logarithmic and exponential equations in a step-by-step process. By PreMath.com.

Learn how to Solve (Challenging) Log Equations with Different Bases in this free math video tutorial by Mario's Math Tutoring.

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

Logarithmic Equations: Very Difficult Problems with Solutions ; x y + y x = l o g 4 32 · +xy=log432 ; x y + y x = 1 2 l o g 2 32 · +xy=21log232 ; x 2 + y 2 x y

Remember that for a logarithm with a positive base the argument (value inside parenthesis) must be positive. Thus, if the solution for x results in a negative

Example 1: Solve the logarithmic equation. log base 2 of x plus 2 plus log base 2 of 3 is equal to. Since we want to transform the left side

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

Solving Logarithmic Equations · log x ( a ) + log x ( b ) = log x ( a b ) \log_x(a) + \log_x(b) = \log_x(ab) logx(a)+logx(b)=logx(ab) · log x ( a ) −

This needs a numeric solution and is in a good form as the right side will change slowly with y. Let y0=0 and iterate. After two iterations we