Solve:
log base 2 (s^2 - 16) = 1 + log base 2 (s - 4)

Question

Solve:
log base 2 (s^2 - 16) = 1 + log base 2 (s - 4)

lgbasallote · Answer

log base 2 (s^2 - 16) -  log base 2 (s - 4) = 1

kropot72 · Answer

Put all the log terms on one side. Can you do that as a first step?

lgbasallote · Answer

log base 2 (s^2 - 16)/(s-4) = 1

log base 2 (s+4)(s-4)/(s-4) = 1

log base 2 (s-4) = 1

change to exponential

2^1 = s - 4

2 = s - 4

2 + 4 = s

6 = s

anonymous · Answer

Hm.. my teacher gave us the answers so we could solve them out, and the answer to this one is no solution? o:

lgbasallote · Answer

oops made a mistake

lgbasallote · Answer

log base 2 (s+4) = 1

not s - 4 sorry

2^1 = s+4

2 - 4 = s

-2 = s

how is that no solution? o.O

anonymous · Answer

Oh because you can't have negatives O: pretty sure

lgbasallote · Answer

maybe...but it doesnt make sense...im thinking your teacher couldve made a mistake :/

anonymous · Answer

All I remember is something about logs and not being allowed to have negatives.. hmm. o;

lgbasallote · Answer

$$\large \log_2 (s^2 - 16) = 1 + \log_2 (s-4)$$ this is the question right??

lgbasallote · Answer

yeah logs cant have negatives

lgbasallote · Answer

oh...oh yeah

anonymous · Answer

Yeahyeah, lol. c:

lgbasallote · Answer

if you sub -2 into $\log_2 (s^2 - 16)$ you'll get a negative number so it's no solution

anonymous · Answer

Ahh, okay. Thank youu C: