log related.........
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log12/log18=a. (Given).
(2log2+log3)/(log2+2log3)=a.
a(log2+2log3)=2log2+log3.
a.log2+2a.log3=2.log2+log3.
(a-2).log2 = (1–2a).log3.
log3/log2=(a-2)./(1–2a)…………..(1).
Now log 16 base24=?
=log16/log24.
=log(2)^4/log 3×(2)^3.
=4.log2 /(log3+3log2) . , divide above and belowby log2.
= 4 / [(log3/log2) +3]
put log3/log2=(a-2)./(1–2a).from eq. (1).
= 4/[(a-2)/(1–2a)+3].
= 4/[(a-2+3–6a)/(1–2a)].
= 4(1–2a)/(1–5a). ,so, option A
(2log2+log3)/(log2+2log3)=a.
a(log2+2log3)=2log2+log3.
a.log2+2a.log3=2.log2+log3.
(a-2).log2 = (1–2a).log3.
log3/log2=(a-2)./(1–2a)…………..(1).
Now log 16 base24=?
=log16/log24.
=log(2)^4/log 3×(2)^3.
=4.log2 /(log3+3log2) . , divide above and belowby log2.
= 4 / [(log3/log2) +3]
put log3/log2=(a-2)./(1–2a).from eq. (1).
= 4/[(a-2)/(1–2a)+3].
= 4/[(a-2+3–6a)/(1–2a)].
= 4(1–2a)/(1–5a). ,so, option A
there is an error in the question..!!
if you solve it acc to question answrr would be 48/14-a and there would be no option available
option is correct man... it's competitive exam question and answer is 1
yes question should be log12 base 18
and not log18 base 12
hummm
thanks a lot
any time..!!
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