Math  /  Algebra

QuestionSolve the equation. Write the solution set with the exact solutions. log4(n4)+log4(n+2)=2\log _{4}(n-4)+\log _{4}(n+2)=2
If there is more than one solution, separate the answers with commas. There is no solution, }\}. The exact solution set is \square

Studdy Solution

STEP 1

What is this asking? We need to find the value(s) of nn that make this logarithmic equation true! Watch out! Remember the logarithm's domain!
We can only take the log of a *positive* number.

STEP 2

1. Combine the logs
2. Rewrite as an exponential equation
3. Solve the quadratic equation
4. Check for validity

STEP 3

We've got two logs added together on one side, so let's use the product rule!
The product rule says logb(x)+logb(y)=logb(xy)\log_b{(x)} + \log_b{(y)} = \log_b{(x \cdot y)}.
This rule is *super* useful for simplifying logarithmic equations.
So, we can rewrite our equation as: log4((n4)(n+2))=2 \log_4((n-4) \cdot (n+2)) = 2

STEP 4

Let's multiply those binomials inside the logarithm: log4(n2+2n4n8)=2 \log_4(n^2 + 2n - 4n - 8) = 2 log4(n22n8)=2 \log_4(n^2 - 2n - 8) = 2

STEP 5

Remember that logb(x)=y\log_b(x) = y means by=xb^y = x.
Here, our base is **4**, the exponent is **2**, and the result is n22n8n^2 - 2n - 8.
So, let's rewrite our equation: 42=n22n8 4^2 = n^2 - 2n - 8

STEP 6

16=n22n8 16 = n^2 - 2n - 8

STEP 7

Let's subtract **16** from both sides to get a nice, clean quadratic equation set to zero: 0=n22n816 0 = n^2 - 2n - 8 - 16 0=n22n24 0 = n^2 - 2n - 24

STEP 8

We're looking for two numbers that multiply to 24-24 and add to 2-2.
Those numbers are 6-6 and 44!
So, we can factor our quadratic like this: 0=(n6)(n+4) 0 = (n - 6)(n + 4)

STEP 9

This gives us two possible solutions for nn: n=6n = 6 and n=4n = -4.

STEP 10

If n=6n = 6, our original equation becomes log4(64)+log4(6+2)=2\log_4(6-4) + \log_4(6+2) = 2, which simplifies to log4(2)+log4(8)=2\log_4(2) + \log_4(8) = 2.
Since 41/2=24^{1/2} = 2 and 43/2=84^{3/2} = 8, we have 12+32=2\frac{1}{2} + \frac{3}{2} = 2, which is 42=2 \frac{4}{2} = 2, so 2=22=2. n=6n=6 works!

STEP 11

If n=4n = -4, we get log4(44)+log4(4+2)=2\log_4(-4-4) + \log_4(-4+2) = 2.
This simplifies to log4(8)+log4(2)=2\log_4(-8) + \log_4(-2) = 2.
Uh oh!
We can't take the logarithm of a negative number!
So, n=4n = -4 is *not* a valid solution.

STEP 12

The exact solution set is {6}\{6\}.

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