Proove that 7 log (16/15) + 5 log (25/4) + 3 log (81/80) = log 2
If ax = by = cz and y2 = xz then prove that logcb = logba
Find the value without using log tables
3 log log37
13 Replies
Akshita
(student)
(108 Points)
Replied 04 June 2012
using the property a*log b = log ba we can write the above as log(16/15)7 + log (25/4)5 + log(81/80)3 and then using the property log a + log b + log c = log a*b*c....
we can write it as log (16/15)7* (25/4)5* (81/80)3 and then solve..
it is equal to log2
Akshita
(student)
(108 Points)
Replied 05 June 2012
since y2 = xz, we can write it as y/x = z/y.
we are given:
ax = by = cz
taking log on the whole equation...
log ax = log by = log cz
using the property log xy = y log x , the above equation can be written as
x log a = y log b = z log c
now we make two cases
x log a = y log b and y log b = z log c
log a / log b = y/x and log b / log c = z/y
logba = y/x and logcb = z/y
in the first line we are given that y/x = z/y
thus we get logba = logcb
Mayank Mall
(CA-Final)
(316 Points)
Replied 05 June 2012
WOW....where did you get this question, I gave my b.com exam today and got this same question in it.....BTW I tried solving it in the paper as Aksh*ta suggested
log a + log b + log c = log a*b*c....
we can write it as log (16/15)7* (25/4)5* (81/80)3 and then solve..
But I did not get the answer this way, later on my friends from maths background told me that to solve this we need to apply another law as
log a/b = log a - log b
Did not tried solving but as they told me that using this we will get a big figure whicih will ultimately be reduced because of the substraction sign and ultimately log 2 will remain.
So trying that i guess i should approach like...
-> ( 7 log 16 - 7 log 15 ) + ( 5 log 25 - 5 log 24 ) + ( 3 log 81 - 3 log 80 )
-> ( 7 log 24 - 7 log 3X5 ) + ( 5 log 52 - 5 log 23X3 ) + ( 3 log 34 - 3 log 24X5 )
-> ( 7 log 24 - ( 7 log 3 + 7 log 5 ) ) + ( 5 log 52 - ( 5 log 23 + 5 log 3 ) ) + ( 3 log 34 - ( 3 log 24 + 3 log 5 ) )
-> ( 7 log 24 - 5 log 23 - 3 log 24 ) + ( - 7 log 3 - 5 log 3 + 3 log 34 ) + ( - 7 log 5 + 5 log 52 - 3 log 5 )
Well believe me from this approach too am getting the wrong answer, am doing something wrong or maybe am missing a point somewhere.....solving the above will take you to the same answer i got today 1 / 24 X 57 X 38 ---- something like this, am just recalling, but the bases are there the powers maybe wrong as I am not solving this damn thing again. I finished my exam way before time but was stuck on this for 45 mins on so and finally gave up, Even i wanna know the answer to this !!!
Tejaswi Kasturi
(student-cpt)
(427 Points)
Replied 05 June 2012
first one the final answer comes to log(2^6 * 3^5)
Mayank Mall
(CA-Final)
(316 Points)
Replied 05 June 2012
@ Tejaswi - log 5 in the denomaintor still exists, how did you cut it......recheck the answer please....is it the same, I'll have to solve it again then :O
Tejaswi Kasturi
(student-cpt)
(427 Points)
Replied 05 June 2012
it doesn't 7(4log2 -log3-log5) + 10(log5 - log2) + (3*(4log3 - 4log2 - log5)) = 28log2 -7log3-7log5+10log5-10log2 +12log3 -12log2-3log5 = 6log2 + 5log3 = log(64*243) I verified it on calculator it is correct
rajiv
(student)
(197 Points)
Replied 05 June 2012
hi friends all the 3 problems are from 9th standard text book published by govt. of Andhra Pradesh.
Mayank Mall
(CA-Final)
(316 Points)
Replied 05 June 2012
hadd....i solved it again.....and got the answer as 1 / 24 X 37 X 55 ........ i donno what to think......and seriously bcom mei itte lower level class 9th ke question aate hai and humse wo bhi ni bante....we planning for CA, hahaha
At Tejaswi.....how did you get this as the middle term 10(log5 - log2)..... it should be 5(2log 5 - 3log 2 - log 3) or (10log 5 - 15log 2 - 5log 3)
Akshita
(student)
(108 Points)
Replied 05 June 2012
i just applied the properties.. and didnt solve it.. but now i did.. and i am getting the same answer as tejaswi....
Tejaswi Kasturi
(student-cpt)
(427 Points)
Replied 05 June 2012
the middle term is log(25/4) not log(25/24)
Tejaswi Kasturi
(student-cpt)
(427 Points)
Replied 05 June 2012
the middle term is log(25/4) not log(25/24)
Mayank Mall
(CA-Final)
(316 Points)
Replied 05 June 2012
yeeeee....gotcha, the question i had was 25/24, rest all was same, and i couldn't solve it.....the question here is 25/4.....so for sure either this or the question in my today's exam was wrong. BTW i am sure the answer you got would be correct now as it aint 25/24 anymore, but my question paper still says that, HUH, wrong question, i should get full marks :D :D :D
Akshita
(student)
(108 Points)
Replied 05 June 2012
i got the answer by taking the middle term as 5 log (25/24)
it is equal to log 2.. provided i am doing the calculations right. which i think i am..
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