An algebra problem by Thevarraaj Chandran

Algebra Level 2

$\log_a x^2y=m$ and $\log_a y/x^3=n$ express $\log_a x/y$ in terms of m and n.

2m+3n -2/5m-3/5n 2 45

3 solutions

Eka Kurniawan
Aug 27, 2014

Parse the equations to $2 log_a x + log_a y = m$ and $log_a y - 3log_a x = n$ . After that, use elimination method to find the value of $log_a x$ and $log_a y$ . Then, substitute those value to $log_a x/y$ which has been changed to $log_a x - log_ay$ . The answer is $\boxed{\frac{-2m-3n}{5}}$

2 nd eqtn is wrong logy - 3logx=n

rahul kumar - 6 years, 9 months ago

thanks for reminding me! but the solution is still the same.

Eka Kurniawan - 6 years, 9 months ago

assume (x/y) = k then m = 3logx - logk
& (-n) = logk+ 2logx eliminate logx then solve for K

Abhinav Verma - 6 years, 9 months ago

i am also getting the same answer bt hw it is -2/5m-3/5n

rahul kumar - 6 years, 9 months ago

what if we consider x/y = (x^2/y)^2/ (y/x^3) and then take log? hence log(x/y) = 2*log (x^2/y) + log(y/x^3) substituting the value of log (x^2/y) and log(y/x^3) it gives log(x/y) = 2m + n. Where am i wrong ??

vivek sedani - 6 years, 8 months ago

I agree to his solution, because that is what i get too

Rio Ri - 6 years, 8 months ago

Thevarraaj Chandran
Aug 21, 2014

log ax^2+log ay=m is 1 and log ay-log ax^3 is 2. Make log ax and log ay as a subject and use subtitution method. From the equation formed, simplify and you should be getting -2/5m-3/5n.

how do you exactly find that logax^2+logay=m is 1?

Sergio Hernandez Diaz - 6 years, 9 months ago

I think the answer is only m/2.

Tejas Joshi - 6 years, 9 months ago

That's what I got too.

Daniel Barnhurst - 6 years, 9 months ago

I didint get this! :(

Rajdip Mahat - 6 years, 9 months ago

good question..

Faraz Mohammed - 6 years, 9 months ago

Sagnik Biswas
Oct 13, 2014

Solution: log(a) x^2 y=m =>log(a) x^2 + log(a) y = m =>2log(a) x+ log(a) y = m ------------- (eq. i)

log(a) y/x^3=n => log(a) y - log(a) x^3 =n =>log(a) y - 3log(a) x =n ----------------- (eq ii)

Subtracting (eq. ii) from (eq. i) we get - 2log(a) x + log(a) y - m + 3log(a) x - log(a) y +n=0 =>5log(a) x = m-n => log(a) x = (m-n)/5

Similarly we get log(a) y = (3m+2n)/5

To find out log(a) x/y i.e. we need to find out log(a) x - log(a) y Therefore, log(a) x - log(a) y = {(m-n)/5} - {(3m+2n)/5} = {m-n-3m-2n}/5 = -2m/5 - 3n/5 (ANS)

An algebra problem by Thevarraaj Chandran

3 solutions

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