Question

1. If log 2, log (2^x-1) and log (2^x+3) are in AP then find the value of x.

Answer 1

Solution:

Since the terms are in AP:

So , 2*log(2^x - 1) = log2 + log(2^x + 3)

or, log(2^x - 1)² = log(2(2^x + 3)) [because log(m.n) = (logm+logn)]

Now, Let 2^x = t
Putting 't' in place of 2^x:

So , (t-1)² = 2(t+3)
or, t² + 1 - 2t = 2t + 6
or, t² - 4t - 5 = 0
or, t^2 - 5t + t - 5 = 0
or, t(t - 5) +1(t - 5) = 0
or, (t -5)(t + 1) = 0

Either:
or, (t - 5) = 0
or, t = 5

Or,
or, (t+1) = 0
or, t = -1

But -1 is negative so 2^x cannot be -1 for any value of x.

So 2^x = 5 ,
x = log₂5 (Answer)