2nd log question clarified re-submitted

Tutoring | Bookmark Us | FREE to Join! New User? Sign Up | Sign In | Help

Students Ask Questions--Tutors Answer
TutorAnswer.com

ask.

answer.

discover.

My Profile

Home > Accounting & Taxes > Open Questions

Doctor Bob

Open Questions

2nd log question clarified re-submitted

It is difficult to submit a complicated fraction at this site.
Please view the problem as

the log sub 2 (sq rt 2) / ( 8)

In other words, the log of 2 times a fraction, the numerator of which is sq rt 2 and the denominator of which is 8. Thank you,.

Posted 198 day ago

Report Abuse

Find Interesting

Email to Friends

Bookmark

Subscribe to Answer Alert

Answers (6)

atta

log(2) * (sqrt(2)/8) => logx * (u/v) = u/vlog(x) = u/v * log(x) = ulog(x) / v. If on the other hand you meant logbase2Of(u/v) = logbase2(u) - logbase2Of(v) thus yielding the following logarithmic expansion:

logbase2Of(sqrt(2)/8) = logbase2Of(sqrt(2)) - logbase2Of(8). This logarithmic expansion has a logarithm of a different base, which thereby entails the change of base formula as in logbaseNOf(x) = logx / logN where both logarithms have their default base of 10. Thus, logbase2Of(sqrt(2)) - logbase2Of(8) = (logsqrt(2) / log2) - (log8 / log2) => log(sqrt(2)) - log(8) / log2 || log(sqrt(2) / 8) / log2 => logu / logv = log(u/v) = logu - logv => log(sqrt(2) / 16).

J.C

Posted 197 day ago

( 0 )

Report Abuse

Doctor Bob

RevDrBob@aol.com

Posted 197 day ago

( 0 )

Report Abuse

YummyCookies

(base 2) [(2^0.5)/8] = x
so: 2^x = [(2^0.5)/8], but 8 = 2^3
so: 2^x = 2^0.5/(2^3) or
(2^x )(2^3) = 2^0.5 or
2^(x+3) = 2^0.5, therefore:
(x+3) = 0.5 or x=-2.5<=Answer.

Posted 198 day ago

( 0 )

Report Abuse

let log(base 2) [(2^0.5)/8] = x
so: 2^x = [(2^0.5)/8], but 8 = 2^3
so: 2^x = 2^0.5/(2^3) or
(2^x )(2^3) = 2^0.5 or
2^(x+3) = 2^0.5, therefore:
(x+3) = 0.5 or x=-2.5<===Answer.

Posted 198 day ago

( 0 )

Report Abuse

matherapist

You may want to use double parenthesis to separate things.
I read your problem as logbase 2( sqrt 2/ 8) = x
then SQRT(2) /8 = 2^x
2^.5=8 x 2^x =2^3 x 2^x = 2^(3+x)
therefore .5=1/2=3+x, x=-5/2

Posted 198 day ago

( 0 )

Report Abuse

MFB13

Hi. Just saw your question around 11:30 AM EST. That written makes more sense to me!

Anyway, here is how you set this up:

log(base2) [ sq rt 2]/8 = X

log(base2) [2^1/2]/2^3 = log(base2) [2^(1/2-6/2)] = log(base2) [2^-5/2] = X

Now we take antilogs of both sides of base 2 of course:

2^-5/2 = 2^x

thus, -5/2 = x

Posted 198 day ago

( 0 )

Report Abuse

Other Open Questions in Accounting & Taxes

» I DON'T HAVE A CLUE- GEOMETRY

» Microsoft Excel 2007

» History Question! Thanks!

» Cell analogy to a school!

» A four digit number

Edit your answer. Click save, when done.
Question Title	2nd log question clarified re-submitted
Your Answer	Character Count ( Max. - 5000 ) : 259