Question

if log2,log(2^x-1) and log (2^x+3) are in A.P then x is equal to? answer=log 5 base 2.

Answer 1

See the attachment for detail solution
Hope it will help you

Answer 2

Answer:

The value of x is  $x=\log_2 5$

Step-by-step explanation:

Given : $\log 2,\log(2^x-1)\text{ and }\log(2^x+3)$ are in A.P.

To find : The x is equal to ?

Solution :

We know that,

If a,b,c are in A.P. then $2b=a+c$

When $\log 2,\log(2^x-1)\text{ and }\log(2^x+3)$ are in A.P.

Then, $2(\log(2^x-1))=\log 2+\log(2^x+3)$

$\log((2^x-1))^2=\log [2(2^x+3)]$

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

Let $2^x=y$

Then, $(y-1)^2=2(y+3)$

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

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

$y=5,-1$

If $2^x=5$

Applying logarithmic property, $a^x=b\Rightarrow x=\log_ba$

$x=\log_2 5$

If $2^x=-1$

Negative value for $2^x$ is not possible.

Therefore, the value of x is  $x=\log_2 5$