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

7log7x

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
log3 3^x = xlog3 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 ?