Question

If 2^a=4^b=12^-c then ab+bc+ca=?

Answer 1

HELLO DEAR,

YOUR QUESTIONS IS--------------3^a = 4^b = 12^(-c)

GIVEN:-

3^a = 4^b = 12^(-c) = k(say)

3^a = k

taking log both sides,

alog3 = logk

a = logk/log3

log3 = logk/a  ---------( 1 )

similarly, 4^b = k

taking log both sides,

blog4 = logk

b = logk/log4

log4 = logk/b -------------( 2 )

AND

12^(-c) = k

taking log both sides,

-clog12 = logk

-clog(3 × 4) = logk

-c{log3 + log4} = logk

from --------- (1) &-------- (2)

= -c{logk/a + logk/b} = logk

= -c{1/a + 1/b} logk = logk

= -c{(a + b/ab)} = 1

= -c{b + a} = ab

= -cb - ac = ab

=  ab + bc + ca = 0

I HOPE ITS HELP YOU DEAR,

THANKS

Answer 2

Note : A small correction is needed. It should be 3^a instead of 2^a.

Given : 3^a = 4^b = 12^(-c).

Let 3^a = 4^b = 12^(-c) = k{Constant].

Then,

(1) 3^a = k

=> log(3^a) = log(k)

=> a log 3 = log k

=> a = log k/log 3.

= > log 3 = (log k/a)   ----- (i)

(2) 4^b = k

=> log(4^b) = log(k)

=> b log 4 = log k

= > b = log k/log 4  

= > log 4 = (log k/b)   ------ (ii)

(3) 12^-c = k

=> log(12^-c) = log(k)

= > -c log 12 = log k

= > -c(log 3 * log 4) = log k [ log ab = log a + log b]

= > -c(log 3 + log 4) = log k

= > -c(log k/a + log k/b) = log k

= > -c(log k(1/a + 1/b)) = log k

= > -c(1/a + 1/b) = 1

= > -c(b + a) = ab

= > -cb - ca = ab

= > ab + bc + ca = 0.

Therefore, the final answer is 0. - Option(2).

Hope this helps!