If `log_10(2),log_10(2^x+1),log_10(2^x-3)` are in A.P then x is equal to (i) `log_2(5)` (ii) `log_2(-1)` (iii) `log_2(1/5)` (iv) `log_5(2)`

Options

A. `log_(2)5`
B. `log_(2)(-1)`
C. `log_(2)(1//5)`
D. `log_(5)2`

Answer

A

Explanation

Since `log_(10)2,log_(10)(2^(x)-1)` and `log_(10)(2^(x)+3)` are in AP.
`:.log_(10)(2^(x)-1)`
`=(log_(10)2+lot_(10)(2^(x)+3))/2`
`implies2log_(10)(2^(x)-1)=log_(10)2(2^(x)+3)`
`rarr log_(10)(2^(x)-1)^(2)=log_(10)2(2^(x)+3)`
`implies(2^(x)-1)^(2)=2(2^(x)+3)`
`implies2^(2x)+1-2^(x+1)=2^(x+1)+6`
`=2^(x)+1-2.2^(x+1)=6`
`implies2^(2x)-4.2^(x)-5=0`
Now let `2^(x)=t`
`impliest^(2)-4t-5=0`
`implies(t-5)(t+1)=0`
`impliest=5` or `t=-1implies2^(x)=5`
[neglecting `2^(x)=-1` as `2^(x)` is always positive]
`impliesx log_(2)2=log_(2)5`
`impliesx=log_(2)5[ :' log_(2)^(2)=1]`