Can anyone help with this logarithm math problem?

Question

This is the problem:

½ log a (x+2) + ½ log a (x-2) = 2/3 log a 27

And, this is what the teacher gave as notes to solve the problem:

Log a √x+2 + log a √x-2 = log a 27^2/3

                                      = log a 9

Cancel out log a -->log a √x+2 √x-2 = log a 9 <--cancel out log a

Which leaves us with: √x+2 √x-2 = 9

So: (√x+2 √x-2)^2 = (9)^2
(x+2) (x-2) = 81
x^2 – 4
√x^2 = √85
x= 9.21

I need help understanding the steps to get the answer. Like, one thing I don’t get is how to get log a 27^2/3 to equal log a 9. Or, why the parentheses "(x+2) (x-2)" turn into square roots in the next step.  

If anyone understands what her math teacher did and how to explain the steps it would be much appreciated.

Answer 1

First, your teacher used the power rule for logs, which allows any coefficient of a log to become the power of the inside. That's how the 2/3, for example, became the power of 27.

Next, the teacher used the product rule of logs to convert the two added logs into the product of their insides, which is how she multiplied √(X+2) * √(X-2) within the left log. On the right, you have 9^(2/3). 

As long as there is only one log per side (and nothing outside of it), you can cancel both of them.

Here I will deviate slightly from your teacher's sequence.

On the left, you have one big square root with a polynomial inside: √(X²-4).

On the right, remember that rational roots in the form X^(a/b) are the same as the the "b" root of X^a. This means you have the cube root of 27-squared:   3√(27²), which also equals (3√27)², because you can take the power or root in either order. The 3√27 = 3, so now you have  3² = 9

Square both sides to clear that remaining square root, and that's how you get:

   √(X²-4) = 9   and then

   X²-4 = 81

   X²    = 85

 

Answer 2

Use these properties to solve the problem:

 

logA + logB = log(AB)

log(A^b) = b logA

c = log_ab ⇒ a^c = b

 

(1/2)log_a(x + 2) + (1/2)log_a(x - 2) = (2/3)log_a27

 

(1/2)log_a[(x + 2)(x  - 2)] = log_a(27^2/3)

 

27^2/3 = (3³)^2/3 = 3² = 9

 

(x + 2)(x - 2) = x² - 4

 

Then

 

(1/2)log_a(x² - 4) = log_a9

 

log_a[(x² - 4)^1/2] = log_a9

 

Now take a⁽⁾ of both sides and get

 

(x² - 4)^1/2 = 9

 

x² - 4 = 9² = 81

 

x² = 81 + 4 = 85

 

x = √85 ≅ 9.2195

 

Note:  The negative square root will not work in the original problem because the argument of the logarithm cannot be negative.