$$\log_{2} (x-2)+\log_{2} (8-x)-\log_{2} (x-5)=3$$

Question

bibby · Answer

Are you aware of the properties of logs?

anonymous · Answer

Product,Quotient,and Power Law right?

bibby · Answer

Mhm. The first 2 are a product, right?

anonymous · Answer

Yup then followed by quotient,then I am stuck.

bibby · Answer

what about the more basic property of logs?$$\log_a(b) = x -> a^x = b$$

anonymous · Answer

Are you suggesting that I solve the left side then use the basic properties of log? to change it's form?

bibby · Answer

I was suggesting something like that but I'm getting unsolvable/unfactorable quadratics so I'm not sure.

kc_kennylau · Answer

$$\log_2(x−2)+\log_2(8−x)−\log_2(x−5)=3\\log_2\frac{(x-2)(8-x)}{x-5}=3\\frac{(x-2)(8-x)}{x-5}=8$$
Then I'm like "what the **** is this ****?!"

alekos · Answer

comes out to a quadratic

x^2 - 2x +30 = 0

alekos · Answer

sorry thats x^2 -2x - 24 = 0

alekos · Answer

just solve for x

anonymous · Answer

Should I write both positive and negative or logarithms are only positive?