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Section 6.4 : Solving Logarithm Equations

In this section we will now take a look at solving logarithmic equations, or equations with logarithms in them. We will be looking at two specific types of equations here. In particular we will look at equations in which every term is a logarithm and we also look at equations in which all but one term in the equation is a logarithm and the term without the logarithm will be a constant. Also, we will be assuming that the logarithms in each equation will have the same base. If there is more than one base in the logarithms in the equation the solution process becomes much more difficult.

Before we get into the solution process we will need to remember that we can only plug positive numbers into a logarithm. This will be important down the road and so we can’t forget that.

Now, let’s start off by looking at equations in which each term is a logarithm and all the bases on the logarithms are the same. In this case we will use the fact that,

In other words, if we’ve got two logs in the problem, one on either side of an equal sign and both with a coefficient of one, then we can just drop the logarithms.

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

With this equation there are only two logarithms in the equation so it’s easy to get on one either side of the equal sign. We will also need to deal with the coefficient in front of the first term.

Now that we’ve got two logarithms with the same base and coefficients of 1 on either side of the equal sign we can drop the logs and solve.

Now, we do need to worry if this solution will produce any negative numbers or zeroes in the logarithms so the next step is to plug this into the original equation and see if it does.

Note that we don’t need to go all the way out with the check here. We just need to make sure that once we plug in the \(x\) we don’t have any negative numbers or zeroes in the logarithms. Since we don’t in this case we have the solution, it is \(x = \frac{1}{5}\).

Okay, in this equation we’ve got three logarithms and we can only have two. So, we saw how to do this kind of work in a set of examples in the previous section so we just need to do the same thing here. It doesn’t really matter how we do this, but since one side already has one logarithm on it we might as well combine the logs on the other side.

Now we’ve got one logarithm on either side of the equal sign, they are the same base and have coefficients of one so we can drop the logarithms and solve.

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

\(x = 6\, :\)

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

\(x = - 2\, :\)

We don’t need to go any farther, there is a logarithm of a negative number in the first term (the others are also negative) and that’s all we need in order to exclude this as a solution.

Be careful here. We are not excluding \(x = - 2\) because it is negative, that’s not the problem. We are excluding it because once we plug it into the original equation we end up with logarithms of negative numbers. It is possible to have negative values of \(x\) be solutions to these problems, so don’t mistake the reason for excluding this value.

Also, along those lines we didn’t take \(x = 6\) as a solution because it was positive, but because it didn’t produce any negative numbers or zero in the logarithms upon substitution. It is possible for positive numbers to not be solutions.

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

We will work this equation in the same manner that we worked the previous one. We’ve got two logarithms on one side so we’ll combine those, drop the logarithms and then solve.

We’ve got two possible solutions to check here.

\(x = 2 :\)

This one is okay.

\(x = 5 :\)

This one is also okay.

In this case both possible solutions, \(x = 2\) and \(x = 5\), end up actually being solutions. There is no reason to expect to always have to throw one of the two out as a solution.

Now we need to take a look at the second kind of logarithmic equation that we’ll be solving here. This equation will have all the terms but one be a logarithm and the one term that doesn’t have a logarithm will be a constant.

In order to solve these kinds of equations we will need to remember the exponential form of the logarithm. Here it is if you don’t remember.

We will be using this conversion to exponential form in all of these equations so it’s important that you can do it. Let’s work some examples so we can see how these kinds of equations can be solved.

To solve these we need to get the equation into exactly the form that this one is in. We need a single log in the equation with a coefficient of one and a constant on the other side of the equal sign. Once we have the equation in this form we simply convert to exponential form.

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.

Now, just as with the first set of examples we need to plug this back into the original equation and see if it will produce negative numbers or zeroes in the logarithms. If it does it can’t be a solution and if it doesn’t then it is a solution.

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

In this case we’ve got two logarithms in the problem so we are going to have to combine them into a single logarithm as we did in the first set of examples. Doing this for this equation gives,

Now, that we’ve got the equation into the proper form we convert to exponential form. Recall as well that we’re dealing with the common logarithm here and so the base is 10.

Here is the exponential form of this equation.

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

\(*x = - 2:\)

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, remember that we don’t exclude a potential solution because it’s negative or include a potential solution because it’s positive. We exclude a potential solution if it produces negative numbers or zeroes in the logarithms upon substituting it into the equation and we include a potential solution if it doesn’t.

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.

\(x = - 4:\)

So, upon substituting this solution in we see that all the numbers in the logarithms are positive and so this IS a solution. Note again that it doesn’t matter that the solution is negative, it just can’t produce negative numbers or zeroes in the logarithms.

In this case, despite the fact that the potential solution is positive we get negative numbers in the logarithms and so it can’t possibly be a solution.

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

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Solving Logarithmic Equations

Generally, there are two types of logarithmic equations. Study each case carefully before you start looking at the worked examples below.

Types of Logarithmic Equations

If you have a single logarithm on each side of the equation having the same base, you can set the arguments equal to each other and then solve. The arguments here are the algebraic expressions represented by \color{blue}M and \color{red}N .

If you have a single logarithm on one side of the equation, you can express it as an exponential equation and solve it .

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.

Since we want to transform the left side into a single logarithmic equation, we should use the Product Rule in reverse to condense it. Here is the rule, just in case you forgot.

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

Let’s check our answer to see if x=7 is a valid solution. Substitute it back into the original logarithmic equation and verify if it yields a true statement.

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

  Example 2: Solve the logarithmic equation.

Start by condensing the log expressions on the left into a single logarithm using the Product Rule. We want to have a single log expression on each side of the equation. Be ready though to solve for a quadratic equation since x will have a power of 2 .

x - 5 = 0 implies that x = 5

x + 2 = 0 implies that x = - 2

So the possible solutions are  x = 5 and  x = - 2 .  Remember to always substitute the possible solutions back to the original log equation.

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

After checking our values of x , we found that x = 5 is definitely a solution. However, x =-2 generates negative numbers inside the parenthesis ( log of zero and negative numbers are undefined) which makes us eliminate x =-2 as part of our solution.

Therefore, the final solution is just \color{blue}x=5 . We disregard x=-2 because it is an extraneous solution.

Example 3: Solve the logarithmic equation.

This is an interesting problem. What we have here are differences of logarithmic expressions on both sides of the equation. Simplify or condense the logs on both sides by using the Quotient Rule.

I will leave it to you to check our potential answers back into the original log equation. You should verify that \color{blue}x=8 is the only solution, while x =-3 is not since it generates a scenario wherein we are trying to get the logarithm of a negative number. Not good!

Example 4: Solve the logarithmic equation.

If you see “log” without an explicit or written base, it is assumed to have a base of 10 . In fact, a logarithm with base 10 is known as the common logarithm .

What we need is to condense or compress both sides of the equation into a single log expression. On the left side, we see a difference of logs which means we apply the Quotient Rule while the right side requires the Product Rule because they’re the sum of logs.

There’s just one thing that you have to pay attention to on the left side. Do you see that coefficient \Large{1 \over 2}\, ?

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

It’s time to check your potential answers. When you check x=0 back into the original logarithmic equation, you’ll end up having an expression that involves getting the logarithm of zero, which is undefined, meaning – not good! So, we should disregard or drop \color{red}x=0 as a solution.

Checking \Large{x = {3 \over 4}} , confirms that indeed \Large{\color{blue}{x = {3 \over 4}}} is the only solution .

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 .

Don’t forget the \pm  symbol.

Check if the potential answers found above are possible answers by substituting them back to the original logarithmic equations.

You should be convinced that the ONLY valid solution is \large{\color{blue}x = {1 \over 2}} which makes \large{\color{red}x = -{1 \over 2}} an extraneous answer.

Example 6: Solve the logarithmic equation.

There is only one logarithmic expression in this equation. We consider this as the second case wherein we have

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

You should verify that the value \color{blue}x=12 is indeed the solution to the logarithmic equation.

Example 7: Solve the logarithmic equation.

Collect all the logarithmic expressions on one side of the equation (keep it on the left) and move the constant to the right side. Use the Quotient Rule to express the difference of logs as fractions inside the parenthesis of the logarithm.

I would solve this equation using the Cross Product Rule. But I have to express first the right side of the equation with the explicit denominator of 1 . That is, 5 = {\large{{5 \over 1}}}

When you check x=1 back to the original equation, you should agree that \large{\color{blue}x=1} is the solution to the log equation.

Example 8: Solve the logarithmic equation.

This problem is very similar to #7. Let’s gather all the logarithmic expressions to the left while keeping the constant on the right side. Since we have the difference of logs, we will utilize the Quotient Rule.

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.

You should agree that \color{blue}x=-32 is the only solution. That makes \color{red}x=4 an extraneous solution, so disregard it.

Example 9: Solve the logarithmic equation

I hope you’re getting the main idea now on how to approach this type of problem. Here we see three log expressions and a constant. Let’s separate the log expressions and the constant on opposite sides of the equation.

It’s obvious that when we plug in x=-8 back into the original equation, it results in a logarithm with a negative number. Therefore, you exclude \color{red}x=-8 as part of your solution.

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

Example 10: Solve the logarithmic equation.

Check this separate lesson if you need a refresher on how to solve different types of Radical Equations .

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:

Condensing Logarithms

Expanding Logarithms

Logarithm Explained

Logarithm Rules


Solving Logarithmic Equations

how to solve difficult logarithmic equations

how to solve difficult logarithmic equations

how to solve difficult logarithmic equations

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Solving Logarithmic Equations

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Equations involving logarithms and unknown variables can often be solved by employing the definition of the logarithm, as well as several of its basic properties :

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

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

\(a\log_x(b) = \log_x(b^a) \)

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

\(x^{\log_x(a)} = a\).

Common mistakes to watch out for include

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

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

The general strategy is to consolidate the logarithms using these properties, and then to take both sides of the equation to the appropriate power in order to eliminate the logarithms if possible.

Solving Logarithmic Equations - Basic

Solving logarithmic equations - intermediate.

For many equations with logarithms, solving them is simply a matter of using the definition of \( \log x \) to eliminate logarithms from the equation and convert it into a polynomial or exponential equation.

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

Another way to view this solution is that we took \( 2 \) to the power of the left side and got \( 3x+1 \), and took \( 2 \) to the power of the right side and got \( 2^4 = 16 \). More complicated logarithmic equations are often simplified by exponentiating both sides.

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

Other equations can be simplified using other properties of logarithms. One difficulty that arises is that eliminating logarithms and solving the resulting equation can introduce spurious solutions. These solutions violate the principle that the argument of the log function must always be positive. It is generally wise to check solutions by plugging them into the original equation and making sure that both sides are defined.

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 \( x=15 \). \(_\square\)

The step in the solution above that was not reversible was the first one: although \( (x-12)(x-6)\) is positive if \( x =3\), \( x-12 \) and \( x-6\) are not themselves positive.

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 \( x \).

More complicated logarithmic equations often involve more than one base. It can help to introduce unknowns to solve for the logarithms first. Another useful identity is \( \log_x(y) = \frac{\log_z(y)}{\log_z(x)} \), especially since \( z\) can be chosen to be whatever simplifies the problem.

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

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

can be written as \( \frac{a}{b} \), where \(a\) and \(b\) are coprime positive integers. What are the last three digits of \(a+b\)?

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 \( x = \frac 12 \). \(_\square\)

What is the sum of all possible real values of \(x\) that satisfies the equation \(x^{\log_{5} x} = \frac{x^3}{25}?\)

Logarithms can also make computation easier in certain practical situations. For example, logarithms with base \(10\) give information about the number of decimal digits in a number.

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. \(_\square\)

Note that this is much easier than multiplying the numbers and comparing them directly. For applications that require large numbers (such as RSA encryption ), this is very important.

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.

Edit: Could give some benefit of doubt to whether the log is base 2 or 10. Any solutions are welcomed.

javabeginner_0101's user avatar

3 Answers 3

Assuming your logs are base $2$ we have $$\lg(2x-24)=2+ \frac 13 \lg8 - \frac 14 x \lg16\\ \lg(x-12)+1=2+1-x\\ lg(x-12)=2-x\\x-12=2^{2-x}\\ 2^x(x-12)=4$$ Clearly $x$ has to be just barely greater than $12$, so let $x=12+y$ Then we have $$y=\frac 1{1024\cdot 2^y}$$ This needs a numeric solution and is in a good form as the right side will change slowly with $y$. Let $y_0=0$ and iterate. After two iterations we have converged to $0.000975902$. Alpha will give you a solution of $y=\frac {W\left(\frac {\log 2}{1024}\right)}{\log 2}$in terms of the Lambert W function where these logs are natural logs.

Added: for base $10$ logs we can do the same. Again $x$ has to be a little greater than $12$ so write $x=12+y$ The equation becomes $$y=\frac {25}{1024\cdot 2^y}$$ which converges to $y\approx 0.0240$

Ross Millikan's user avatar

This is indeed a very good introduction to Lambert function.

Sooner or later, you will learn that any equation which can write or rewrite $$A+Bx+C\log(D+Ex)=0$$ has solution(s) which espress(es) in terms of this function.

In the case of natural logarithms, using the steps shown in Ross Millikan's answer, you would end with $$x=12+\frac{1}{\log_e (2)}W\left(\frac{e^2 \log_e (2)}{4096}\right)$$

Assuming logarithms in base $2$ as the numbers suggest, then, as Ross Millikan answered, $$x=12+\frac{1}{\log_e (2)}W\left(\frac{\log_e (2)}{1024}\right)$$

Assuming logarithms in base $10$, $$x=12+\frac{1}{\log_e (2)}W\left(\frac{25 \log_e (2)}{1024}\right)$$

Now, since the argument is quite small, you can approximate the value of $W(t)$ using the expansion $$W(t)=t-t^2+\frac{3 }{2}t^3-\frac{8 }{3}t^4+O\left(t^5\right)$$ or, better, using Padé approximants such as $$W(t)=\frac{t }{1+t }$$ $$W(t)=\frac{t+\frac{4}{3} t^2}{1+\frac{7 }{3}t+\frac{5 }{6}t^2 }$$

Claude Leibovici's user avatar

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

$x-12$ = $100$ * $(2^{-x})$ ,

x = 100*(2^(-x)) + 12 ,

On my calculator, to get an approximation, i just started with some random number ( $1$ in my case), and entered $100$ *( $2^{-ans}$ ) + $12$ , where ans is the previous value. By repeatedly evaluating this, it converges to to the value of $x$ = $12.0240111$ . In theory, this should work from any initial value.

Samuel Price's user avatar

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