# How do you solve [log_4(x+3)] + [log_4(2-x)] = 1?

Nov 29, 2015

$x = 1 , - 2$

#### Explanation:

Now toy need to know some basic rules to attempt this;

No1.

# $\implies {a}^{x} = b$

No2.

${\log}_{a} b + {\log}_{a} c = {\log}_{a} b c$

Now we are good to go

$\left[{\log}_{4} \left(x + 3\right)\right] + \left[{\log}_{4} \left(2 - x\right)\right] = 1$

${\log}_{4} \left(x + 3\right) + {\log}_{4} \left(2 - x\right) = 1$

Now lets Identify our common base$\implies 4$

Now

Using rule number 2 Lets combine

$\implies {\log}_{4} \left(x + 3\right) \left(2 - x\right) = 1$

Now lets go back to our first result

Take a look

${\log}_{a} b = x$

Our base here is 4 and our b is $\left(x + 3\right) \left(2 - x\right)$

So lets wrap it up;

${4}^{1} = \left(x + 3\right) \left(2 - x\right)$

$4 = \left(x + 3\right) \left(2 - x\right)$

Lets now distribute and multiply;

$\left(x + 3\right) \left(2 - x\right) = \left(2\right) \left(x\right) + \left(- x\right) \left(x\right) + \left(3\right) \left(2\right) + \left(3\right) \left(- x\right)$

$\left(x + 3\right) \left(2 - x\right) = 2 x - {x}^{2} + 6 - 3 x$

$\left(x + 3\right) \left(2 - x\right) = - {x}^{2} + 6 - x$

=> $- {x}^{2} + 6 - x = 4$

# =>$- {x}^{2} - x + 2 = 0$

Now lets try factoring

So

$a \cdot b = - 2$, $a + b = - 1$

So the only possibility to factor is splitting the middle term in pars of
1 and -2

=>$- {x}^{2} - 2 x + x + 2 = 0$

$- x \left(x + 2\right) + 1 \left(x + 2\right) = 0$

$\implies \left(1 - x\right) \left(x + 2\right) = 0$

=> There are 2 roots;
$x = 1 , - 2$

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Check
=

$\implies {\log}_{4} \left(x + 3\right) \left(2 - x\right) = 1$
......................................................................................................................
x = 1

$\implies {\log}_{4} \left(1 + 3\right) \left(2 - 1\right) = 1$
$\implies {\log}_{4} 4 = 1$

Which is right ; So it is a valid solution
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x =- 2

$\implies {\log}_{4} \left(- 2 + 3\right) \left(2 - \left(- 2\right)\right) = 1$
$\implies {\log}_{4} \left(1\right) \left(2 + 2\right) = 1$

$\implies {\log}_{4} 4 = 1$

Which is right ; So it is a valid solution

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