The question is to evaluate each logarithm:

Question

My professor wants us to show work. Could you please tell me if I'm on the right track? Thank you!!!

1. log6^(1/36)= -2

2. ln(1/e^4)= 1/e^2

3. log7^1= 0

4. log4^64= ln64/ln4= 3

5. log10^8= ln8/ln10= .9031

6. log3^(radical 3)= I wasn't sure how to show the work on this one. I plugged it into my calculator and got .8263

Answer 1

#1
How did you get -2 ??

log 6^(1/36)
= (1/36) log 6 = appr .0216

#2
ln(1/e^4)
= ln1 - ln e^4
= 0 - 4lne
= 0 - 4(1) = -4

#3
log 7^1
= log 7 = appr .845

#4
log 4^64
= 64 log4 = appr 38.532

#5
log 10^8
= 8log10
= 8

#6 --- the only one you got right

Answer 2

Some of these blokes use

log6^(1/36) to mean log₆(1/36), which is indeed -2

#2 ln(1/e^4) = ln(e^-4) = -4
#3 ok
#4 ok
#5 since the other logs specify a base, this must be just a common log, and log 10^8 = 8
#6 log₃√3 = log₃3^(1/2) = 1/2

Answer 3

Hi thank you so much for helping me. Yeah, I am sorry. I do not know how to make the base numbers small like that. Thank you both again for the help!

Answer 4

Thanks Steve
Had an inkling that's what he was doing, but then it was contradicted in how he worked the last one.
btw, was it overkill in my solution to
http://www.jiskha.com/display.cgi?id=1461944692

Almost gave up on it, but found some clues on a webpage.

Answer 5

The subscripts are a trick of HTML. Most of the codes do not work on this site, but the sub and /sub tags work.

For superscripts, the sup and /sup tags work.

enclose the words in <> brackets to make them into HTML tags.