General Aptitude

Question

If $log_32,log_3(2^a-5),log_3(2^a-\frac{7}{2})$ are in AP, then ‘a’ equals

(a)2

(b)4

(c)5

(d)None of these

Comments

2 ??

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Use basic property of log and arithmetic series.

$log_3(2^a-5) - log_32 = log_3(2^a- \frac{7}{2})-log_3(2^a-5)$.

you will get a = 2.

Answer 1

Three terms a,b,c are in AP then

2*b = a + c.

$2*log_{3} (2^{x}-5) = log_{3}2 + log_{3}(2^{x} - 7/2)$

$(2^{x}-5)^{2} = 2*(2^{x}-7/2)$

$2^{2x} - 12*2^{x} + 32.$

Here $2^{x} = y.$

Then $y^{2} - 12y + 32$

$y = 4$

$y = 8.$

$2^{x} = 4 = 2^{2} ==>> x = 2$

$2^{x} = 8 = 2^{3} ==>> x = 3.$

Comments

Yes, I was also getting 2 as answer.Wanted to confirm.Thanks.