# Thread: Logarithmic equations

1. ## Logarithmic equations

Question: Solve teh logarithmic equation. Be sure to reject any value that is not in the domain of the original logarithmic expressions. Give the exact answers

(Log 2^(x+2)) + (log 2^(x-4))=4

I know the answer is 6, but is completely clueless on how to get 6.

here is what my work looks like:

= log 2^(x+2)(x-4)=4

= log 2 ^ (x^2-2x-8)=4

= log 2^(x^2-2x-12)

= x=2 +/- square root of 52 / 2

that is where i stopped, because 52 is not a perfect square.

Test is tomorrow!

2. Originally Posted by dto
Question: Solve teh logarithmic equation. Be sure to reject any value that is not in the domain of the original logarithmic expressions. Give the exact answers

(Log 2^(x+2)) + (log 2^(x-4))=4

I know the answer is 6, but is completely clueless on how to get 6.

here is what my work looks like:

= log 2^(x+2)(x-4)=4

= log 2 ^ (x^2-2x-8)=4

= log 2^(x^2-2x-12)

= x=2 +/- square root of 52 / 2

that is where i stopped, because 52 is not a perfect square.

Test is tomorrow!
I believe you made a mistake with the laws of exponents.

Also I don't understand your notation. Do you mean $\log_2 (x+2)$ or $\log_{10}\, (2^{x+2})$

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Edit I'm going to assume the former because $f(6) = 4$ for it.

$\log_2 (x+2) + \log_2 (x-4) = 4$

$\log_2 [(x+2)(x-4)] = \log_2 (x^2-2x-8) = 4$

$x^2-2x-8 = 2^4 = 16$

$x^2-2x-24=0$

Solve the quadratic - this one will factorise