(Untitled)

[abdicate_it]

Okay. Just two more questions.

How would I solve:

log2 16

2 would be the base and 16 would be the end result?
so it would be log2=16?
and then from there?
Orrr....?

And for
2logx - log(x+1) = log4 - log3
The answer is 2.
How is the answer two?

Thank you soo much for helping.

Comments

[lyesya]

log 16 / log 2
change of base theorem (or something like that)

2 log x - log(x+1) = log 4/3
2 log x = log 4(x+1) / 3
log x = log 4(x+1) / 3 divided by 2
so x = 4(x+1) / 6
6x=4x+4
2x=4
x=2

[bob]

is that the log-base-two of sixteen?

[happyevilslosh]

Asking someone to find the solution for log216 is the same as trying to solve 2x=16 in terms of x. This one is actually fairly easy if you know your powers of 2. You can also use the identity (if I get my numbers round the right way - you'll want to double check this) logxy=logzy / logzx.

The other has been solved above.

[happyevilslosh]

P.S. The Boondock Saints is a friggin' awesome movie! :)

[pilot_student]

For the second there's also a property of logs that says:
AlogB = log(B^A)