log 2 , log(2^x - 1), log(2^x + 3 ) ..... find f

Since the above is an arthimatical progression, then:

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

But we know that log a - log b = log a/b

==> log...

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log 2 , log(2^x - 1), log(2^x + 3 ) ..... find f

Since the above is an arthimatical progression, then:

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

But we know that log a - log b = log a/b

==> log (2^x -1)/2 = log (2^x+3)/(2^x -1)

==> (2^x -1)/2 = (2^x + 3)/(2^x -1)

Cross multiply:

==? (2^x -1)^2 = 2(2^x +3)

Expand brackets:

==> 2^2x -2*2^x + 1 = 2*2^x + 6

==> (2^x)^2 - 4*2^x -5 =0

Let y = 2^x

==> y^2 - 4y -5 = 0

==> (y-5)(y+1) = 0

==> y= 5 ==> 5= 2^x ==> x = log 2 (5)

==? y= -1 ==> -1 = 2^x (impossible)

Then the solution is :

x = log 2 (5)