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Math  /  Algebra

QuestionSolve the following logarithmic equation. log⁡2(x+11)−2=−log⁡2(x+8)\log _{2}(x+11)-2=-\log _{2}(x+8)

Studdy Solution

STEP 1

What is this asking? We need to find the value(s) of xx that make this logarithmic equation true! Watch out! Remember logarithmic properties and be careful about potential extraneous solutions – always check your answers!

STEP 2

1. Rewrite the equation
2. Combine the logarithms
3. Convert to exponential form
4. Solve the quadratic equation
5. Check for extraneous solutions

STEP 3

Let's **move** all the logarithmic terms to one side of the equation.
We can do this by adding log⁡2(x+8)\log_2(x+8) to both sides: log⁡2(x+11)+log⁡2(x+8)−2=0 \log_2(x+11) + \log_2(x+8) - 2 = 0

STEP 4

Now, let's **isolate** the logarithmic terms by adding 22 to both sides: log⁡2(x+11)+log⁡2(x+8)=2 \log_2(x+11) + \log_2(x+8) = 2

STEP 5

Using the **product property of logarithms**, log⁡b(m)+log⁡b(n)=log⁡b(m⋅n)\log_b(m) + \log_b(n) = \log_b(m \cdot n), we can combine the two logarithms on the left side: log⁡2((x+11)⋅(x+8))=2 \log_2((x+11) \cdot (x+8)) = 2 log⁡2(x2+19x+88)=2 \log_2(x^2 + 19x + 88) = 2

STEP 6

Remember that log⁡b(a)=c\log_b(a) = c means bc=ab^c = a.
So, we can rewrite our logarithmic equation in exponential form: 22=x2+19x+88 2^2 = x^2 + 19x + 88 4=x2+19x+88 4 = x^2 + 19x + 88

STEP 7

Let's **set the equation to zero** by subtracting 44 from both sides: 0=x2+19x+84 0 = x^2 + 19x + 84

STEP 8

Now, we can **factor** the quadratic: 0=(x+7)(x+12) 0 = (x+7)(x+12)

STEP 9

This gives us two potential solutions: x=−7x = \mathbf{-7} and x=−12x = \mathbf{-12}.

STEP 10

Remember, the argument of a logarithm must be **positive**.
Let's check x=−7x = \mathbf{-7}: log⁡2(−7+11)−2=−log⁡2(−7+8) \log_2(-7+11) - 2 = -\log_2(-7+8) log⁡2(4)−2=−log⁡2(1) \log_2(4) - 2 = -\log_2(1) 2−2=−0 2 - 2 = -0 0=0 0 = 0 So, x=−7x = \mathbf{-7} is a valid solution!

STEP 11

Now let's check x=−12x = \mathbf{-12}: log⁡2(−12+11)−2=−log⁡2(−12+8) \log_2(-12+11) - 2 = -\log_2(-12+8) log⁡2(−1)−2=−log⁡2(−4) \log_2(-1) - 2 = -\log_2(-4) Uh oh!
We have **negative arguments** inside the logarithms.
This means x=−12x = \mathbf{-12} is an extraneous solution and we must discard it.

STEP 12

The solution to the equation is x=−7x = \mathbf{-7}.

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