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Posted by **Chopsticks** on Monday, February 23, 2009 at 4:56pm.

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?

- Math+Log -
**Reiny**, Monday, February 23, 2009 at 5:06pmis it

7^{log7x}

if so, then the answer is x

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

- Math+Log -
**Chopsticks**, Monday, February 23, 2009 at 5:07pmAhh yea it looks like that. Hey can you teach me how to do all those subscripts and those impossible math symbols?

- Math+Log -
**Chopsticks**, Monday, February 23, 2009 at 5:08pm27log27x = x

13log13x=x

??

- Math+Log -
**Reiny**, Monday, February 23, 2009 at 5:14pmcorrect, 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 -
**Chopsticks**, Monday, February 23, 2009 at 5:15pmlog3(3^x)

The first 3 is a subscript

Would that = x?

Also with

log15(15^x) = x ?

log221(221^x)= x?

- Math+Log -
**Damon**, Monday, February 23, 2009 at 5:29pmloga(a^x) = x

for any old a

because

loga (a^x) = x loga(a)

but loga(a) = 1

- Math+Log -
**Reiny**, Monday, February 23, 2009 at 5:29pmremember that log (a^n) = n log a with any base

so

log_{3}3^x = xlog_{3}3

= x

yes you are right.

- Math+Log -
**Chopsticks**, Monday, February 23, 2009 at 5:30pmohh 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 -
**Chopsticks**, Monday, February 23, 2009 at 5:33pmWait so is it x or 1?

- Math+Log -
**Reiny**, Monday, February 23, 2009 at 5:42pmit 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 -
**Chopsticks**, Monday, February 23, 2009 at 5:45pmoh 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

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