Homework Help: Math+Log

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

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?

Math+Log - Reiny, Monday, February 23, 2009 at 5:06pm
is 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 5:07pm
Ahh 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 5:08pm
27log27x = x
13log13x=x

??
Math+Log - Reiny, Monday, February 23, 2009 at 5:14pm
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 - Chopsticks, Monday, February 23, 2009 at 5:15pm
log3(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 5:29pm
loga(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 5:29pm
remember that log (a^n) = n log a with any base
so
log₃ 3^x = xlog₃ 3
= x

yes you are right.
Math+Log - Chopsticks, Monday, February 23, 2009 at 5:30pm
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 - Chopsticks, Monday, February 23, 2009 at 5:33pm
Wait so is it x or 1?
Math+Log - Reiny, Monday, February 23, 2009 at 5:42pm
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 - Chopsticks, Monday, February 23, 2009 at 5:45pm
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

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