is it
7log7x
if so, then the answer is x
basic log rule: a^(log k, with base a) = k
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?
11 answers
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
??
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)
(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?
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
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.
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
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 ?
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
I thought it was 15x at first, but then i realize that it was its suppose to be 15^x
Thanks