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

Lesson Explainer: Logarithmic Equations with Like Bases Mathematics

In this explainer, we will learn how to solve logarithmic equations with like bases using the laws of exponents and logarithms.

A logarithmic equation is an equation with an unknown variable in part of the logarithm, usually the argument. In the case of logarithmic equations with a single logarithm, it is possible to rewrite them in exponential form to make them easier to solve. Let’s recall the definition of a logarithm to show how to do this.

Definition: Logarithms

For an exponential equation in the form 𝑎=𝑛, where 𝑎>0, this can be written as the logarithmic equation log𝑛=𝑥, where 𝑎 is the base of the logarithm, 𝑛 is the argument, and 𝑥 is the exponent.

Therefore, 𝑎=𝑛𝑛=𝑥log.

We can see from the definition above that if an unknown variable is in the argument, 𝑛, then by rewriting log𝑛=𝑥 as 𝑎=𝑛 we can make 𝑛 the subject of the equation, making it possible to solve for an unknown variable.

In the first example, we will discuss how to find an unknown variable in the argument by rewriting the logarithm in its exponential form, as detailed above.

Example 1: Solving Logarithmic Equations with a Single Logarithm

Given that log(𝑣+3)=1, find the value of 𝑣.

Answer

As we are given a logarithmic equation with a single logarithm, we can rearrange the logarithm to make the argument, 𝑣+3, the subject by rewriting the logarithm as an exponential equation. Recall that log𝑛=𝑥𝑎=𝑛, where 𝑎>0 and 𝑛>0.

Therefore, for the logarithmic equation log(𝑣+3)=1, we can rewrite this as 12=𝑣+3.

Evaluating 12 to the power of 1, and solving for 𝑣, we get 12=𝑣+312=𝑣+3𝑣=123=9.

Therefore, the value of 𝑣 in the logarithmic equation log(𝑣+3)=1 is 9.

In the next example, we will solve another logarithmic equation with a single logarithm by rewriting it as an exponential equation, but this time with an unknown in the base.

Example 2: Solving a Logarithmic Equation with a Single Logarithm

What is the solution set of the equation log64=2?

Answer

As we are given a logarithmic equation with a single logarithm, with an unknown variable in the base, by rewriting this as an exponential equation, we can more easily solve it. Recall that log𝑛=𝑥𝑎=𝑛, where 𝑎>0 and 𝑛>0.

Therefore, we can rearrange log64=2 by writing it as the exponential equation (𝑥+2)=64.

We can solve this by taking the square root of both sides of the equation, giving us 𝑥+2=±8.

As the base, 𝑥+2, must be greater than zero, +8 is the only possible solution, so we discard 8. Solving for 𝑥, we then get 𝑥+2=8𝑥=82=6.

Therefore, the solution set for log64=2 is {6}.

So far, we have considered logarithmic equations with a single logarithm. Next, we will discuss how to solve logarithmic equations with multiple logarithms of the same base.

Let’s recall the laws of logarithms for like bases and special values.

Law: Laws of Logarithms for Like Bases and Special Values

For a logarithm with base 𝑎, where 𝑎>0,

  • log𝑎=1;
  • log1=0;
  • multiplication law: logloglog𝑥𝑦=𝑥+𝑦, where 𝑥>0 and 𝑦>0;
  • division law: logloglog𝑥𝑦=𝑥𝑦, where 𝑥>0 and 𝑦>0;
  • power law: loglog𝑥=𝑛𝑥, where 𝑥>0.

Where we have like bases, we can use the laws of logarithms to first combine logarithms, and then solve by either rearranging or making the arguments equal, depending on the equation.

We will demonstrate how to use the laws of logarithms for like bases to solve a logarithmic equation in our next example.

Example 3: Finding the Solution Set for a Logarithmic Equation with Like Bases

Determine the solution set of the equation logloglog(𝑥6)+(𝑥+6)=64 in .

Answer

In the logarithmic equation logloglog(𝑥6)+(𝑥+6)=64, as all parts of the equation have logarithms with the same base, 8, we can apply the laws of logarithms for like bases in order to simplify and solve.

Since we are adding two logarithms on the left-hand side of the equation, we can use the law for multiplication of logarithms to simplify this. The law states that logloglog𝑥𝑦=𝑥+𝑦, where 𝑎>0, 𝑥>0, and 𝑦>0.

Applying this to the left-hand side, we get logloglogloglog(𝑥6)+(𝑥+6)=64(𝑥6)(𝑥+6)=64.

Now, since both the left-hand side and right-hand side of the equation are written as a single logarithm with the same base, therefore the arguments must be equal, giving us (𝑥6)(𝑥+6)=64.

Expanding the brackets and rearranging to equal zero, we get 𝑥6𝑥+6𝑥36=64𝑥100=0.

Factoring and solving, we then get 𝑥100=0(𝑥10)(𝑥+10)=0𝑥=10𝑥=10.or

Now, since the argument of a logarithm must be greater than zero, both 𝑥6 and 𝑥+6 must be greater than zero. As such, 10 is not a valid solution, so 𝑥 must equal 10.

Therefore, the solution set of the equation logloglog(𝑥6)+(𝑥+6)=64 is {10}.

In the next example, we will consider how to use multiple laws of logarithms with like bases to solve a logarithmic equation.

Example 4: Finding the Solution Set for a Logarithmic Equation with Like Bases

Find the solution set of 4(𝑥+5)(𝑥3)=625logloglog in .

Answer

In the logarithmic equation 4(𝑥+5)(𝑥3)=625logloglog, as all parts of the equation have logarithms with the same base, 2, we can apply the laws of logarithms for like bases in order to simplify and solve.

Since we have a multiplication of a number by a logarithm in the first term, we need to use the power law for logarithms, which states that loglog𝑥=𝑛𝑥, where 𝑎>0 and 𝑥>0.

Therefore, 4(𝑥+5)=(𝑥+5)loglog, giving us 4(𝑥+5)(𝑥3)=625(𝑥+5)(𝑥3)=625.loglogloglogloglog

Next, as we are subtracting one logarithm from another with the same base on the left-hand side of the equation, we can simplify this using the division law for logarithms, which states that logloglog𝑥𝑦=𝑥𝑦, where 𝑎>0, 𝑥>0, and 𝑦>0.

Applying this law, we get logloglogloglog(𝑥+5)(𝑥3)=625(𝑥+5)(𝑥3)=625.

Now, since both the left-hand side and the right-hand side of the equation are written as a single logarithm with the same base, the arguments must be equal, giving us 𝑥+5𝑥3=625.

As both the numerator and the denominator of the left-hand side are a power of 4, we can take the positive and negative fourth root, giving us 𝑥+5𝑥3=±625=±5.

Solving for +5, we get 𝑥+5𝑥3=5𝑥+5=5(𝑥3)𝑥+5=5𝑥154𝑥=20𝑥=5.

Solving for 5, we get 𝑥+5𝑥3=5𝑥+5=5(𝑥3)𝑥+5=5𝑥+156𝑥=10𝑥=53.

Therefore, the solution set of 4(𝑥+5)(𝑥3)=625logloglog is 53,5.

In our last example, we will discuss how to use the laws of logarithms to solve a geometric problem.

Example 5: Using Laws of Logarithms to Solve a Geometric Problem

Given that 𝐴𝐵𝐴𝐶 and 𝐴𝐷𝐵𝐶, determine the value of 𝑥. Give your answer to the nearest tenth.

Answer

As seen in the figure, we have one large right triangle, 𝐴𝐵𝐶, with side 𝐴𝐵 and hypotenuse 𝐵𝐶=𝐵𝐷+𝐷𝐶, with 𝐵𝐷 and 𝐷𝐶 given, as well as a smaller right triangle, 𝐴𝐵𝐷, inside the larger one, with side 𝐵𝐷 and hypotenuse 𝐴𝐵 given. As both triangles are right and share a common angle, 𝐵, they must be similar triangles. Therefore, the ratios of the sides are equal, giving us 𝐵𝐶𝐴𝐵=𝐴𝐵𝐵𝐷, which is the same as [𝐴𝐵]=𝐵𝐶×𝐵𝐷.

We can see from the diagram that 𝐴𝐵=83,𝐵𝐷=1024=2=102,𝐷𝐶=32=2=52,loglogloglogloglog and 𝐵𝐶=𝐵𝐷+𝐷𝐶=102+52=152.logloglog

Therefore, substituting into [𝐴𝐵]=𝐵𝐶×𝐵𝐷, we get 83=102152192=15022=1921502=32252=±32252=±425.logloglogloglogloglog

As we are working with lengths of sides of triangles, log2 must be positive, so log2=425.

To find 𝑥, we can rewrite this as an exponential equation since log𝑛=𝑥𝑎=𝑛, where 𝑎>0 and 𝑛>0.

Therefore, log2=425𝑥=2.

To solve for 𝑥, we recall that if 𝑥=𝑦, then 𝑥=𝑦, giving us 𝑥=2𝑥=2𝑥=2𝑥=1.8.tothenearest10th

Therefore, 𝑥 is 1.8 to the nearest tenth.

In this explainer, we have discussed how to solve logarithmic equations using their exponential forms, and using laws of logarithms for equations with like bases. Let’s recap the key points.

Key Points

  • A logarithmic equation is an equation with an unknown in part of the logarithm.
  • If a logarithmic equation has an unknown in its base or argument and contains a single logarithm, then we can rearrange it using its exponential form.
  • If a logarithmic equation contains multiple logarithms with the same base, then we can use the laws of logarithms to simplify it, and then either equate arguments if there are single logarithms on both sides or write it in its exponential form if there is a logarithm on only one side.

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