Math / Algebra

QuestionSolve the logarithmic equation. Be sure to reject any value of $x$ that is not in the domain of the original logarithmic expressions. Give the exact answer. $\log _{6}(x+8)+\log _{6}(x+3)=1$
Solve the equation. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The solution set is $\square$ \}. (Simplify your answer. Use a comma to separate answers as needed.) B. There are infinitely many solutions. C. There is no solution.

Studdy Solution

STEP 1

What is this asking? We need to find the value(s) of $x$ that satisfy the given logarithmic equation, keeping in mind that we can't take the logarithm of a negative number or zero! Watch out! We need to make sure our solution(s) don't make the arguments of the logarithms negative or zero.
Sometimes, perfectly good-looking solutions can cause trouble, so we gotta check 'em!

STEP 2

1. Combine the logarithms
2. Rewrite the logarithmic equation as an exponential equation
3. Solve the quadratic equation
4. Check for extraneous solutions

STEP 3

We've got $\log_{6}(x+8) + \log_{6}(x+3) = 1$ .
Since we're adding logarithms with the same base, we can multiply their arguments.
This is because of the product rule of logarithms: $\log_b(m) + \log_b(n) = \log_b(m \cdot n)$ .

STEP 4

Applying this rule, we get $\log_{6}((x+8) \cdot (x+3)) = 1$ , which simplifies to $\log_{6}(x^2 + 11x + 24) = 1$ .
Awesome!

STEP 5

Remember that logarithms and exponentials are inverse operations.
So, we can rewrite our logarithmic equation $\log_{6}(x^2 + 11x + 24) = 1$ in exponential form.

STEP 6

The base is 6, the exponent is 1, and the result is $x^2 + 11x + 24$ .
This gives us the equation $6^1 = x^2 + 11x + 24$ , or simply $6 = x^2 + 11x + 24$ .
Looking good!

STEP 7

To solve $6 = x^2 + 11x + 24$ , let's subtract 6 from both sides to get $0 = x^2 + 11x + 18$ .
Now we have a nice quadratic equation!

STEP 8

We can factor this quadratic as $(x+2)(x+9) = 0$ .
This gives us two possible solutions: $x+2=0$ or $x+9=0$ .

STEP 9

Solving these gives us $x = -2$ and $x = -9$ .
Almost there!

STEP 10

Remember, we can't have negative arguments inside our logarithms.
Let's check $x = -2$ .
Substituting into the original equation, we get $\log_{6}(-2+8) + \log_{6}(-2+3) = 1$ , which simplifies to $\log_{6}(6) + \log_{6}(1) = 1$ .
Since $\log_{6}(6) = 1$ and $\log_{6}(1) = 0$ , we have $1 + 0 = 1$ , which is true!
So, $x = -2$ is a valid solution.

STEP 11

Now let's check $x = -9$ .
Substituting, we get $\log_{6}(-9+8) + \log_{6}(-9+3) = 1$ , which simplifies to $\log_{6}(-1) + \log_{6}(-6) = 1$ .
Uh oh!
We can't take the logarithm of negative numbers.
So, $x = -9$ is an extraneous solution and we must reject it.

STEP 12

The solution set is $\{-2\}$ .
We choose A, and fill in the box with -2.

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Studdy Solution

STEP 1

STEP 2

1. Combine the logarithms
2. Rewrite the logarithmic equation as an exponential equation
3. Solve the quadratic equation
4. Check for extraneous solutions

STEP 3

We've got $\log_{6}(x+8) + \log_{6}(x+3) = 1$ .
Since we're adding logarithms with the same base, we can multiply their arguments.
This is because of the product rule of logarithms: $\log_b(m) + \log_b(n) = \log_b(m \cdot n)$ .

STEP 4

Applying this rule, we get $\log_{6}((x+8) \cdot (x+3)) = 1$ , which simplifies to $\log_{6}(x^2 + 11x + 24) = 1$ .
Awesome!

STEP 5

Remember that logarithms and exponentials are inverse operations.
So, we can rewrite our logarithmic equation $\log_{6}(x^2 + 11x + 24) = 1$ in exponential form.

STEP 6

The base is 6, the exponent is 1, and the result is $x^2 + 11x + 24$ .
This gives us the equation $6^1 = x^2 + 11x + 24$ , or simply $6 = x^2 + 11x + 24$ .
Looking good!

STEP 7

To solve $6 = x^2 + 11x + 24$ , let's subtract 6 from both sides to get $0 = x^2 + 11x + 18$ .
Now we have a nice quadratic equation!

STEP 8

We can factor this quadratic as $(x+2)(x+9) = 0$ .
This gives us two possible solutions: $x+2=0$ or $x+9=0$ .

STEP 9

Solving these gives us $x = -2$ and $x = -9$ .
Almost there!

STEP 10

STEP 11

Now let's check $x = -9$ .
Substituting, we get $\log_{6}(-9+8) + \log_{6}(-9+3) = 1$ , which simplifies to $\log_{6}(-1) + \log_{6}(-6) = 1$ .
Uh oh!
We can't take the logarithm of negative numbers.
So, $x = -9$ is an extraneous solution and we must reject it.

STEP 12

The solution set is $\{-2\}$ .
We choose A, and fill in the box with -2.

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Studdy Solution

STEP 1

STEP 2

1. Combine the logarithms2. Rewrite the logarithmic equation as an exponential equation3. Solve the quadratic equation4. Check for extraneous solutions

STEP 3

STEP 4

Applying this rule, we get log⁡6((x+8)⋅(x+3))=1\log_{6}((x+8) \cdot (x+3)) = 1log6​((x+8)⋅(x+3))=1, which simplifies to log⁡6(x2+11x+24)=1\log_{6}(x^2 + 11x + 24) = 1log6​(x2+11x+24)=1.Awesome!

STEP 5

Remember that logarithms and exponentials are **inverse operations**.So, we can rewrite our logarithmic equation log⁡6(x2+11x+24)=1\log_{6}(x^2 + 11x + 24) = 1log6​(x2+11x+24)=1 in exponential form.

STEP 6

The base is **6**, the exponent is **1**, and the result is x2+11x+24x^2 + 11x + 24x2+11x+24.This gives us the equation 61=x2+11x+246^1 = x^2 + 11x + 2461=x2+11x+24, or simply 6=x2+11x+246 = x^2 + 11x + 246=x2+11x+24.Looking good!

STEP 7

To solve 6=x2+11x+246 = x^2 + 11x + 246=x2+11x+24, let's subtract 6 from both sides to get 0=x2+11x+180 = x^2 + 11x + 180=x2+11x+18.Now we have a nice quadratic equation!

STEP 8

We can **factor** this quadratic as (x+2)(x+9)=0(x+2)(x+9) = 0(x+2)(x+9)=0.This gives us two possible solutions: x+2=0x+2=0x+2=0 or x+9=0x+9=0x+9=0.

STEP 9

Solving these gives us x=−2x = -2x=−2 and x=−9x = -9x=−9.Almost there!

STEP 10

STEP 11

STEP 12

The solution set is {−2}\{-2\}{−2}.We choose A, and fill in the box with -2.

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1. Combine the logarithms
2. Rewrite the logarithmic equation as an exponential equation
3. Solve the quadratic equation
4. Check for extraneous solutions

Applying this rule, we get $\log_{6}((x+8) \cdot (x+3)) = 1$ , which simplifies to $\log_{6}(x^2 + 11x + 24) = 1$ .
Awesome!

Remember that logarithms and exponentials are inverse operations.
So, we can rewrite our logarithmic equation $\log_{6}(x^2 + 11x + 24) = 1$ in exponential form.

The base is 6, the exponent is 1, and the result is $x^2 + 11x + 24$ .
This gives us the equation $6^1 = x^2 + 11x + 24$ , or simply $6 = x^2 + 11x + 24$ .
Looking good!

To solve $6 = x^2 + 11x + 24$ , let's subtract 6 from both sides to get $0 = x^2 + 11x + 18$ .
Now we have a nice quadratic equation!

We can factor this quadratic as $(x+2)(x+9) = 0$ .
This gives us two possible solutions: $x+2=0$ or $x+9=0$ .

Solving these gives us $x = -2$ and $x = -9$ .
Almost there!

The solution set is $\{-2\}$ .
We choose A, and fill in the box with -2.