# Math+Log

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Simplify the expression:
The first 7 is regular size and the log7x are small (the second 7 is subscripted only).
7log7x =

How do you work this?

• Math+Log -

is it

7log7x

if so, then the answer is x

basic log rule: a^(log k, with base a) = k

• Math+Log -

Ahh yea it looks like that. Hey can you teach me how to do all those subscripts and those impossible math symbols?

• Math+Log -

27log27x = x
13log13x=x

??

• Math+Log -

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)

• Math+Log -

log3(3^x)

The first 3 is a subscript

Would that = x?

Also with

log15(15^x) = x ?

log221(221^x)= x?

• Math+Log -

loga(a^x) = x
for any old a
because
loga (a^x) = x loga(a)
but loga(a) = 1

• Math+Log -

remember that log (a^n) = n log a with any base
so
log3 3^x = xlog3 3
= x

yes you are right.

• Math+Log -

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

• Math+Log -

Wait so is it x or 1?

• Math+Log -

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 ?

• Math+Log -

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