Question #82e26

"A" is rather similar to "4" so I suppose there is a typo in between.
Putting `4` instead "A" it works. So

`log_2x+log_2(2x+7)=log_2 2^2 = 2`

`log_2(x(2x+7))=log_2 4 equiv x(2x+7)=4`

Solving for `x`

`x = -4` and `x = 1/2`

`log_2x+log_2(2x+7)=log_2A` .......(1)
Solve:
`log_2x+log_2(2x+7)=2` .......(2)

We see from inspection that LHS of both equations (1) and (2) are equal. Therefore, RHS of both the equations must be equal.
`=> log_2A=2`
Writing this in exponential form by definition of `log` we get
`2^2=A`,
or `A=4`
Inserting the value of `A` in (1), equation (2) becomes
`log_2x+log_2(2x+7)=log_2 4`
LHS can be written using the rule for, addition of `log` functions as
`log_2x+log_2(2x+7)=log_2 [x(2x+7)]`
or `log_2 (2x^2+7x)`
Equating with RHS
`log_2 (2x^2+7x)=log_2 4`
Taking `"anti" log_2` of both sides, we get
`2x^2+7x=4`
or `2x^2+7x-4=0`,

Using the split the middle term method
`2x^2+8x-x-4=0`
or `2x(x+4)-(x+4)=0`
or `(x+4)(2x-1)=0`
We get two roots by putting each factor `=0`

1. `x+4=0`
   or `x=-4`
2. `2x-1=0`
   or `x=1/2`

Since `log` of a `-ve` number is not defined, therefore, only valid solution is `x=1/2`

The derivations for x = -4 and 1/2 are as in other two answers.

`x-4` is inadmissible.

log of negative argument is unreal.

`log_2 (-4)=log_2(4i^2)`

`=log_2 4+log_2(i^2)`

`=.log_2(2^2)+2 log_2(i^2)`

`=2(log_2 2+log_2 i)`

The RHS `log_2 A` is a hint to solve for the RHS in the second equation. Of course, the second is sufficient, without the first..