Question #82e26

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Question #82e26

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Answer 1 · 2016-05-30T11:27:12

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

Explanation:

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

${log}_{2} x + {log}_{2} (2 x + 7) = {log}_{2} 2^{2} = 2$ $log_2x+log_2(2x+7)=log_2 2^2 = 2$

${log}_{2} (x (2 x + 7)) = {log}_{2} 4 \equiv x (2 x + 7) = 4$ $log_2(x(2x+7))=log_2 4 equiv x(2x+7)=4$

Solving for $x$ $x$

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

Answer 2 · 2016-05-30T11:36:46

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

Explanation:

${log}_{2} x + {log}_{2} (2 x + 7) = {log}_{2} A$ $log_2x+log_2(2x+7)=log_2A$ .......(1)
Solve:
${log}_{2} x + {log}_{2} (2 x + 7) = 2$ $log_2x+log_2(2x+7)=2$ .......(2)

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

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

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

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

Answer 3 · 2016-05-30T12:19:45

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

Explanation:

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

$x - 4$ $x-4$ is inadmissible.

log of negative argument is unreal.

${log}_{2} (- 4) = {log}_{2} (4 i^{2})$ $log_2 (-4)=log_2(4i^2)$

$= {log}_{2} 4 + {log}_{2} (i^{2})$ $=log_2 4+log_2(i^2)$

$= . {log}_{2} (2^{2}) + 2 {log}_{2} (i^{2})$ $=.log_2(2^2)+2 log_2(i^2)$

$= 2 ({log}_{2} 2 + {log}_{2} i)$ $=2(log_2 2+log_2 i)$

The RHS ${log}_{2} A$ $log_2 A$ is a hint to solve for the RHS in the second equation. Of course, the second is sufficient, without the first..

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Question #82e26

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