My Math Forum help me with this exponential real quick please
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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: (5-3ln(7)) - (2-3ln(4)) shouldn't this equal 3-3ln(4/7)? the answer that i should get is 3-3ln(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^(y-x) thus, y - x = ln(7/4) (5 - 3y) - (2 - 3x) 5 - 2 - 3 (y - x) 3 - 3 ln(7/4) Thanks from The
 April 25th, 2014, 10:39 AM #5 Newbie   Joined: Apr 2014 From: London Posts: 13 Thanks: 1 Thanks a lot mate! Thanks from MATHEMATICIAN
 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 (5-3ln(7))-(2-3ln(4))= 5-3ln(7)-2+3ln(4)= 5-2+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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