Math+Log
posted by Chopsticks on .
Simplify the expression:
The first 7 is regular size and the log7x are small (the second 7 is subscripted only).
7log7x =
would the answer be 1?
How do you work this?

is it
7^{log7x}
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
13log13x=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) 
log3(3^x)
The first 3 is a subscript
Would that = x?
Also with
log15(15^x) = x ?
log221(221^x)= x? 
loga(a^x) = x
for any old a
because
loga (a^x) = x loga(a)
but loga(a) = 1 
remember that log (a^n) = n log a with any base
so
log_{3} 3^x = xlog_{3} 3
= x
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
Thanks 
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
Thanks