LIVE GST Certification Course by CA Arun Chhajer begins 21st April. Register Now!!

Qt - logarithms

rajiv (student) (197 Points)

04 June 2012

Proove that 7 log (16/15) + 5 log (25/4) + 3 log (81/80) = log 2

If a^x = b^y = c^z and y² = xz then prove that log_cb = log_ba

Find the value without using log tables

3 log log₃7

Reply
Follow
Share

13 Replies

Akshita (student) (108 Points)
Replied 04 June 2012

using the property a*log b = log b^a we can write the above as log(16/15)⁷ + log (25/4)⁵ + log(81/80)³and 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)³and 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:

a^x = b^y = c^z

taking log on the whole equation...

log a^x = log b^y = log c^z

using the property log x^y = 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

log_ba = y/x and log_cb = z/y

in the first line we are given that y/x = z/y

thus we get log_ba = log_cb

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)³and 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 / 2⁴X 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