Logarithm bases -> continue

[HJ:)]

Hey, my teacher told me to put 1/c + 1/d into a single fraction.
He made it sound so easy, but when I came to doing it, I got little confused

And also he told me to express log (base ab)x only in terms of logs.

Can you help?? ASAP

[SharkSpider]

1/c + 1/d = (c + d)/cd

logab(x) = lnx/lnab

At the risk of sounding a bit harsh, you should probably be at the point where this stuff comes naturally, and if not, then I'd reccomend just doing more practice questions from logs and algebra.

[FAI_faiii]

isnt it cd/(c+d)?
cuz for example, log(base4)64,log(base 8)64, log(base32)64
log(base4)64=3,so c=3, log(base 8)64=2, so d=2
log(base32)64=1.2 which is log(base ab)X=1.2
if it is (c+d)/cd, the answer should be (3+2)/3*2=5/6
so i think it should be cd/(c+d), so 3*2/(3+2)=1.2
tell me if im right or wrong

[SharkSpider]

1/c + 1/d = c/cd + d/cd = (c+d)/cd

1/1 + 1/1 = (1+1)/1*1 = 2, NOT 1*1/(1+1) = 1/2.

[HJ:)]

[quote name='SharkSpider' post='28284' date='Nov 12 2008, 07:43 AM']1/c + 1/d = c/cd + d/cd = (c+d)/cd

1/1 + 1/1 = (1+1)/1*1 = 2, NOT 1*1/(1+1) = 1/2.[/quote]

hy, thanks for helping out, but I asked my teacher and he said it was cd/(c+d)

Edited November 12, 2008 by HJ:)

[Alfabeta]

[quote name='HJ:)' post='28300' date='Nov 12 2008, 07:26 PM']hy, thanks for helping out, but I asked my teacher and he said it was cd/(c+d) [/quote]


Your teacher shall not tell you such things;) though you're right;) (CD)/c+d..... god damen ****in' hell you will not even dear to think of the proof for this statement:O:O