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April 25th, 2014, 10:18 AM  #1 
Newbie Joined: Apr 2014 From: London Posts: 13 Thanks: 1  help me with this exponential real quick please
alright query: (53ln(7))  (23ln(4)) shouldn't this equal 33ln(4/7)? the answer that i should get is 33ln(7/4) can someone explain why exactly this is..? thanks i may be missing something really big here 
April 25th, 2014, 10:25 AM  #2 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 901 Thanks: 61 Math Focus: सामान्य गणित 
3  3 ln(7/4) is correct 
April 25th, 2014, 10:36 AM  #3 
Newbie Joined: Apr 2014 From: London Posts: 13 Thanks: 1  Not exacly the answer i wanted
it is correct yes, but shouldn't 3ln4  3ln7 be 3ln(4/7)? Since it is the 4 minus the 7?

April 25th, 2014, 10:36 AM  #4 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 901 Thanks: 61 Math Focus: सामान्य गणित 
let y = ln 7 , x = ln 4 or, e^y = 7, e^x = 4 then, 7/4 = (e^y)/(e^x) = e^(yx) thus, y  x = ln(7/4) (5  3y)  (2  3x) 5  2  3 (y  x) 3  3 ln(7/4) 
April 25th, 2014, 10:39 AM  #5 
Newbie Joined: Apr 2014 From: London Posts: 13 Thanks: 1 
Thanks a lot mate!

April 25th, 2014, 10:40 AM  #6 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 901 Thanks: 61 Math Focus: सामान्य गणित 
yes your question has 4 minus 7 but your answer also has minus infront of ln its like taking minus common and having  ( 7 minus 4) 
April 25th, 2014, 10:40 AM  #7 
Newbie Joined: Apr 2014 From: Albania Posts: 8 Thanks: 2 
(53ln(7))(23ln(4))= 53ln(7)2+3ln(4)= 52+3ln(4)3ln(7)= 3 + 3(ln(4)ln(7))= 3+3ln(4/7); log(n)ln(m)=ln(n/m) 
April 25th, 2014, 10:43 AM  #8 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 901 Thanks: 61 Math Focus: सामान्य गणित 
THE thanks for THANKS your thanks is the first thanks that i've earned in mmf 

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