HJ:) 2 Report post Posted November 11, 2008 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 Share this post Link to post Share on other sites
SharkSpider 35 Report post Posted November 11, 2008 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. Share this post Link to post Share on other sites
FAI_faiii 0 Report post Posted November 11, 2008 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 Share this post Link to post Share on other sites
SharkSpider 35 Report post Posted November 12, 2008 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. Share this post Link to post Share on other sites
HJ:) 2 Report post Posted November 12, 2008 (edited) [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:) Share this post Link to post Share on other sites
Alfabeta 7 Report post Posted November 12, 2008 [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 Share this post Link to post Share on other sites