if so, then the answer is x
basic log rule: a^(log k, with base a) = k
Ahh yea it looks like that. Hey can you teach me how to do all those subscripts and those impossible math symbols?
27log27x = x
correct, if they have the same appearance as your first question.
(As tutors some have been given special privileges that students don't have.
for example I can post a link which normally does not work here)
The first 3 is a subscript
Would that = x?
log15(15^x) = x ?
loga(a^x) = x
for any old a
loga (a^x) = x loga(a)
but loga(a) = 1
remember that log (a^n) = n log a with any base
log3 3^x = xlog3 3
yes you are right.
ohh yea ok thanks.
so the subscript is like the answer to 15^x? and the only way to make it = 15 is if x = 1
Wait so is it x or 1?
it is x
look at Damon's last two lines:
" loga (a^x) = x loga(a)
but loga(a) = 1 "
so wouldn't loga (a^x) = x(1) = x ?
oh yea it is x
I thought it was 15x at first, but then i realize that it was its suppose to be 15^x