Lesson Explainer: Logarithmic Equations with Different Bases | Nagwa Lesson Explainer: Logarithmic Equations with Different Bases | Nagwa

Lesson Explainer: Logarithmic Equations with Different Bases Mathematics

In this explainer, we will learn how to solve logarithmic equations involving logarithms with different bases.

Let’s first recall the relationship between logarithmic and exponential forms.

Definition: Relationship between Logarithmic and Exponential Forms

For 𝑥>0 and base 𝑎>0, 𝑎1, the logarithmic form 𝑦=𝑥log is equivalent to the exponential form 𝑥=𝑎, which allows us to convert from one form to another once we identify 𝑎, 𝑥, and 𝑦.

An exponential function 𝑦=𝑎 is the inverse of the logarithmic function 𝑦=𝑥log. This means if you raise 𝑎 to the power of log of 𝑥 with base 𝑎 or if you raise 𝑎 to the power of 𝑥 first then take the log with base 𝑎 of the result, you get back 𝑥: 𝑎=(𝑎)=𝑥.loglog

The relationship between logarithmic and exponential forms allows us to deduce the properties satisfied by the logarithmic forms, known as the laws of logarithms, that follow from the laws of exponents. Let’s recall the laws of logarithms.

Definition: Laws of Logarithms

Suppose 𝑥, 𝑦, 𝑎, and 𝑏 are positive numbers with 𝑎,𝑏1. The laws of logarithms are

  • product: logloglog(𝑥𝑦)=𝑥+𝑦,
  • division: logloglog𝑥𝑦=𝑥𝑦,
  • powers: loglog(𝑥)=𝑛𝑥,
  • change of base: logloglog𝑥=𝑥𝑎.

These will be useful to solving logarithmic equations with different bases, especially the last one, which allows us to convert a logarithm of one base to another. In order to see where this comes from, first note that 𝑥=𝑎,𝑎=𝑏,loglog which follow directly from the fact that exponents and logarithms are inverses. If we substitute the second expression into the first, we obtain 𝑥=𝑏.loglog

Using the law of exponents (𝑝)=𝑝, we can rewrite this as 𝑥=𝑏.loglog

Taking the logarithm of both sides with base 𝑏, we find loglogloglogloglog𝑥=𝑎𝑥𝑥=𝑥𝑎, as required. This formula along with the laws of logarithms and the equivalent exponential form allow us to solve logarithmic equations involving logarithms of different bases. Another fact that will be important for solving logarithmic equations is loglog𝑥=𝑦𝑥=𝑦, which follows directly from the relationship between logarithmic and exponential forms, in particular, from the fact that the logarithmic and exponential functions are strictly monotonic functions. We can show this directly using the laws of logarithms: logloglogloglog𝑥=𝑦𝑥𝑦=0𝑥𝑦=0.

Converting the last expression to exponential form, we find 𝑥𝑦=𝑎=1, and thus, 𝑥=𝑦, as required.

As an example, suppose we want to find the solutions to the logarithmic equation loglog(2𝑥1)=𝑥.

We first convert the logarithm on the left-hand side to a logarithm of base 2 using the change in base formula: loglogloglogloglog(2𝑥1)=(2𝑥1)4=(2𝑥1)2=12(2𝑥1), where we have used the fact that log𝑎=𝑛 to obtain the last line. Thus, using this and the laws of logarithms, the given logarithmic equation becomes 12(2𝑥1)=𝑥(2𝑥1)=2𝑥(2𝑥1)=𝑥.loglogloglogloglog

Using the fact that we established, if loglog𝑥=𝑦, then 𝑥=𝑦, or converting to exponential form, we obtain 2𝑥1=𝑥𝑥2𝑥+1=0(𝑥1)=0.

Thus, the only solution to the logarithmic equation loglog(2𝑥1)=𝑥 is 𝑥=1.

Now, let’s consider a few examples to practice and deepen our understanding of solving logarithmic equations. In the first example, we have two logarithms of different bases and the unknown 𝑥 appears inside the logarithm.

Example 1: Finding the Solution Set of a Logarithmic Equation over the Set of Real Numbers

Find the solution set of loglog𝑥=4 in .

Answer

In this example, we want to determine the solution set of a particular logarithmic equation with different bases and the unknown appearing inside the logarithm.

In order to solve the equation, we will make use of the change of base formula, logloglog𝑥=𝑥𝑎, and the power law, 𝑛𝑥=(𝑥).loglog

On applying the change of base formula, we can write the right-hand side of the given logarithmic equation, log4, as a logarithm of base 3 using 9=3 and log𝑎=𝑛: loglogloglogloglog4=49=43=124.

Using this, the given equation becomes loglogloglogloglog𝑥=124𝑥=4𝑥=2.

Since loglog𝑥=𝑥𝑥=𝑥, we have 𝑥=2. Thus, the solution set is given by {2}.

Now, let’s consider an example where a logarithmic equation contains two logarithms of different bases and an unknown appearing in each.

Example 2: Finding the Solution Set of a Logarithmic Equation over the Set of Real Numbers

Determine the solution set of the equation loglog𝑥+𝑥+3=0 in .

Answer

In this example, we want to determine the solution set of a particular logarithmic equation with different bases and the unknown appearing inside two logarithms of different bases.

In order to solve the equation, we will make use of the change of base formula, logloglog𝑥=𝑥𝑎, and the power law, 𝑛𝑥=(𝑥).loglog

On applying the change of base formula, we can write the second term on the left-hand side of the given logarithmic equation, log𝑥, as a logarithm of base 3 using 243=3 and log𝑎=𝑛: logloglogloglogloglog𝑥=𝑥243=𝑥3=15𝑥=𝑥.

On substituting this into the given logarithmic equation, we obtain loglogloglogloglog𝑥+𝑥+3=0𝑥+𝑥+3=02𝑥=3𝑥=32.

For 𝑥>0 and base 𝑎>0, 𝑎1, the logarithmic form 𝑦=𝑥log is equivalent to the exponential form 𝑥=𝑎.

Finally, converting log𝑥=32 to exponential form we have 𝑥=3=13=13×3=133.

Thus, the solution set is 133.

In the next example, we will find the solution to a logarithmic equation that contains the sum of three logarithms of different bases and the unknown appearing inside each of the logarithms.

Example 3: Finding the Solution Set of a Logarithmic Equation over the Set of Real Numbers

Find the solution set of logloglog𝑥+𝑥+𝑥=21 in .

Answer

In this example, we want to determine the solution set of a particular logarithmic equation with different bases and the unknown appearing inside the logarithm.

In order to solve the equation, we will make use of the change of base formula, logloglog𝑥=𝑥𝑎, and the power law, 𝑛𝑥=(𝑥).loglog

On applying the change of base formula, we can write the second term on the left-hand side of the given logarithmic equation, log𝑥, as a logarithm of base 2 using 4=2 and log𝑎=𝑛, loglogloglogloglog𝑥=𝑥4=𝑥2=12𝑥, and similarly, for the third term, log𝑥, using 16=2 as loglogloglogloglog𝑥=𝑥16=𝑥2=14𝑥.

Substituting these expressions into the logarithmic equation, we obtain loglogloglogloglogloglog𝑥+𝑥+𝑥=21𝑥+12𝑥+14𝑥=2174𝑥=21𝑥=12.

For 𝑥>0 and base 𝑎>0, 𝑎1, the logarithmic form 𝑦=𝑥log is equivalent to the exponential form 𝑥=𝑎.

Thus, converting log𝑥=12 to exponential form, we obtain 𝑥=2=4096.

Therefore, the solution set is {4096}.

Now, let’s consider an example where we have to find the solution to a logarithmic equation that contains the sum of the reciprocal of three logarithms of different bases and the unknown appearing inside each of the logarithms.

Example 4: Finding the Solution Set of a Logarithmic Equation over the Set of Real Numbers

Find the solution set of 1𝑥+1𝑥+1𝑥=3logloglog in .

Answer

In this example, we want to determine the solution set of a particular logarithmic equation with different bases and the unknown appearing inside three logarithms of different bases.

In order to solve the equation, we will make use of the change of base formula, loglogloglogloglog𝑥=𝑥𝑎1𝑥=𝑎𝑥, and the power law, 𝑛𝑥=(𝑥).loglog

On applying the change of base formula, we can write the second term on the left-hand side of the given logarithmic equation, 1𝑥log, as a logarithm of base 2 using 4=2 and log𝑎=𝑛, 1𝑥=4𝑥=2𝑥=2𝑥,loglogloglogloglog and similarly, for the third term, 1𝑥log, using 8=2 as 1𝑥=8𝑥=2𝑥=3𝑥.loglogloglogloglog

Substituting these expressions into the logarithmic equation, we obtain 1𝑥+1𝑥+1𝑥=31𝑥+2𝑥+3𝑥=36𝑥=3𝑥=2.loglogloglogloglogloglog

For 𝑥>0 and base 𝑎>0, 𝑎1, the logarithmic form 𝑦=𝑥log is equivalent to the exponential form 𝑥=𝑎.

Thus, converting log𝑥=2 to exponential form we find 𝑥=2=4.

Thus, the solution set is {4}.

In the next example, we will solve a logarithmic equation that contains logarithms of different bases, including a fractional base, and an unknown appearing inside the logs as a linear and quadratic term.

Example 5: Solving Logarithmic Equations Involving Laws of Logarithms

Find the solution set of loglog𝑥𝑥=6 in .

Answer

In this example, we want to determine the solution set of a particular logarithmic equation with different bases and the unknown appearing inside two logarithms of different bases.

In order to solve the equation, we will make use of the change of base formula, logloglog𝑥=𝑥𝑎, and the power law, 𝑛𝑥=(𝑥).loglog

On applying the change of base formula, we can write the second term on the left-hand side of the given logarithmic equation, log𝑥, as a logarithm of base 3 using 13=3 and log𝑎=𝑛: logloglogloglogloglog𝑥=𝑥=𝑥3=𝑥=2𝑥.

Finally, substituting this expression into the given logarithmic equation, we have loglogloglogloglog𝑥𝑥=6𝑥+2𝑥=63𝑥=6𝑥=2.

For 𝑥>0 and base 𝑎>0, 𝑎1, the logarithmic form 𝑦=𝑥log is equivalent to the exponential form 𝑥=𝑎.

Finally, converting log𝑥=2 to exponential form, we have 𝑥=3=9.

Thus, the solution set is {9}.

As we have seen in the previous examples, the change of base rule for logarithms also allows us to evaluate expressions of the form log𝑥 by rewriting this logarithm of base 𝑎 as a logarithm of base 𝑎 (i.e., 𝑏=𝑎) along with the power law loglog(𝑥)=𝑛𝑥 and the fact that log𝑎=𝑘: loglogloglog𝑥=𝑥𝑎=1𝑘𝑥.

Now, let’s consider an example where we have a logarithmic equation that contains a logarithm of the logarithm with two different bases and an unknown appearing inside the logarithm as a quadratic.

Example 6: Solving Logarithmic Equations Involving Laws of Logarithms and Quadratic Equations

Solve loglog𝑥8𝑥=1, where 𝑥.

Answer

In this example, we want to determine the solution of a particular logarithmic equation with different bases and an unknown appearing inside the logarithm of a logarithm in the form of a quadratic equation.

For 𝑥>0 and base 𝑎>0, 𝑎1, the exponential form 𝑦=𝑎 is equivalent to the logarithmic form 𝑥=𝑦log, which allows us to convert from one form to another once we identify 𝑎, 𝑥, and 𝑦.

Converting loglog𝑥8𝑥=1 to exponential form, we obtain log𝑥8𝑥=2=2.

Repeating the process, we obtain 𝑥8𝑥=3𝑥8𝑥9=0(𝑥9)(𝑥+1)=0.

Therefore, we obtain 𝑥=1 or 𝑥=9. The solution set is {1,9}.

So far, the examples we have considered had the unknown variable 𝑥, which we have to solve for, appearing inside the logarithm itself. We can also find solutions to logarithmic equations where the unknown can appear in the base of the logarithm.

In the next example, let’s consider a logarithmic equation with a triple logarithm of different bases and an unknown appearing as the base.

Example 7: Solving Logarithmic Equations over the Set of Real Numbers

Solve logloglog36=0, where 𝑥.

Answer

In this example, we want to determine the solution of a particular logarithmic equation with three different bases and an unknown appearing as a base of a logarithm.

Recall that, for 𝑥>0 and base 𝑎>0, 𝑎1, the logarithmic form 𝑦=𝑥log is equivalent to the exponential form 𝑥=𝑎. The common logarithm where no base is specified has base 10: loglog=.

Converting logloglog36=0 to exponential form, we obtain loglog36=10=1.

Repeating this process for the resulting expression, we get log36=2=2.

And repeating it lastly again, we obtain 𝑥=36.

Thus, the solutions are 𝑥=6 or 𝑥=6, but we ignore the second solution since the logarithm of a negative base (i.e., log36), is undefined.

Therefore, the solution to the logarithmic equation is 𝑥=6.

Now, let’s consider an example where we have a logarithmic equation containing logarithms of different bases and an unknown appearing both inside the logarithm and the base of the same logarithm.

Example 8: Finding the Solution Set of Exponential Equations Involving Logarithms over the Set of Real Numbers

Find the solution set of 𝑥=10loglog in .

Answer

In this example, we want to determine the solution of a particular logarithmic equation with two different bases and an unknown appearing inside and appearing as a base of a logarithm.

In order to solve the given equation, we will make use of the power law: 𝑛𝑥=(𝑥).loglog

On applying this and using log𝑎=1, we obtain loglog𝑥=6𝑥=6 and loglog10=6410=64.

Substituting these into the given equation, we obtain 𝑥=64.

The solutions are 𝑥=2 and 𝑥=2; however, we will ignore the second as the logarithm of a negative number or with a negative base is undefined. The solution set is therefore {2}.

Using the change of base formula, we can interchange the argument of the logarithm and the base. Changing the logarithm log𝑥 to base 𝑏=𝑥 and using log𝑥=1, we have loglogloglog𝑥=𝑥𝑎=1𝑎.

Suppose we want to find the solution of loglog5=3; we want to determine the values of 𝑥 that satisfy this logarithmic equation. Since log𝑎=1, this simplifies to log5=1.

Thus, converting this to exponential form with base 𝑥, we obtain 𝑥=𝑥=5.

Finally, let’s consider an example where the unknown 𝑥 that we have to solve for appears both inside the logarithm and as a base of another logarithm.

Example 9: Finding the Solution Set of Logarithmic Equations over the Set of Real Numbers

Determine the solution set of the equation loglog𝑥+254=10 in .

Answer

In this example, we want to determine the solution of a particular logarithmic equation with two different bases and an unknown appearing inside and appearing as a base of a logarithm.

In order to solve the logarithmic equation, we will make use of the change in base formula: logloglog𝑥=𝑥𝑎.

If we use the base 𝑏=𝑥, this formula allows us to interchange the argument of the logarithm and the base: loglogloglog𝑥=𝑥𝑎=1𝑎, where we used the fact that log𝑥=1. Using this, the given equation becomes loglog𝑥+25𝑥=10.

So, if we let 𝑦=𝑥log, then we have to solve 𝑦+251𝑦=10.

Multiplying both sides of this equation by 𝑦 and rearranging, we obtain 𝑦10𝑦+25=0(𝑦5)=0.

Thus, 𝑦=𝑥=5.log

For 𝑥>0 and base 𝑎>0, 𝑎1, the logarithmic form 𝑦=𝑥log is equivalent to the exponential form 𝑥=𝑎.

Thus, upon converting to exponential form, we have 𝑥=4=1024.

Thus, the solution set is {1024}.

Let’s summarize what has been learned in this explainer.

Key Points

  • In order to solve the logarithmic equations, we made use of the laws of logarithms:
    • product: logloglog(𝑥𝑦)=𝑥+𝑦,
    • division: logloglog𝑥𝑦=𝑥𝑦,
    • powers: loglog(𝑥)=𝑛𝑥,
    • change of base: logloglog𝑥=𝑥𝑎.
  • If we have a logarithm of base 𝑎, we can also convert this to a logarithm of base 𝑎 using loglog𝑥=1𝑘𝑥.
  • We can also interchange the base and the argument of a logarithm using loglog𝑥=1𝑎.
  • We can also solve logarithmic equations by converting the logarithmic form 𝑦=𝑥log to the exponential form 𝑥=𝑎.

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