# Question #82e26

May 30, 2016

$x = - 4$ and $x = \frac{1}{2}$

#### Explanation:

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

${\log}_{2} x + {\log}_{2} \left(2 x + 7\right) = {\log}_{2} {2}^{2} = 2$

${\log}_{2} \left(x \left(2 x + 7\right)\right) = {\log}_{2} 4 \equiv x \left(2 x + 7\right) = 4$

Solving for $x$

$x = - 4$ and $x = \frac{1}{2}$

May 30, 2016

Here we go and obtain
$x = \frac{1}{2}$

#### Explanation:

${\log}_{2} x + {\log}_{2} \left(2 x + 7\right) = {\log}_{2} A$ .......(1)
Solve:
${\log}_{2} x + {\log}_{2} \left(2 x + 7\right) = 2$ .......(2)

We see from inspection that LHS of both equations (1) and (2) are equal. Therefore, RHS of both the equations must be equal.
$\implies {\log}_{2} A = 2$
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
${\log}_{2} x + {\log}_{2} \left(2 x + 7\right) = {\log}_{2} 4$
LHS can be written using the rule for, addition of $\log$ functions as
${\log}_{2} x + {\log}_{2} \left(2 x + 7\right) = {\log}_{2} \left[x \left(2 x + 7\right)\right]$
or ${\log}_{2} \left(2 {x}^{2} + 7 x\right)$
Equating with RHS
${\log}_{2} \left(2 {x}^{2} + 7 x\right) = {\log}_{2} 4$
Taking $\text{anti} {\log}_{2}$ of both sides, we get
$2 {x}^{2} + 7 x = 4$
or $2 {x}^{2} + 7 x - 4 = 0$,

Using the split the middle term method
$2 {x}^{2} + 8 x - x - 4 = 0$
or $2 x \left(x + 4\right) - \left(x + 4\right) = 0$
or $\left(x + 4\right) \left(2 x - 1\right) = 0$
We get two roots by putting each factor $= 0$

1. $x + 4 = 0$
or $x = - 4$
2. $2 x - 1 = 0$
or $x = \frac{1}{2}$

Since $\log$ of a $- v e$ number is not defined, therefore, only valid solution is $x = \frac{1}{2}$

May 30, 2016

A=4 and $x = \frac{1}{2}$

#### Explanation:

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

$x - 4$ is inadmissible.

log of negative argument is unreal.

${\log}_{2} \left(- 4\right) = {\log}_{2} \left(4 {i}^{2}\right)$

$= {\log}_{2} 4 + {\log}_{2} \left({i}^{2}\right)$

$= . {\log}_{2} \left({2}^{2}\right) + 2 {\log}_{2} \left({i}^{2}\right)$

$= 2 \left({\log}_{2} 2 + {\log}_{2} i\right)$

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..