Question

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

Answer 1

`x = 1 ,-2`

Explanation:

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

No1.

`log_a b = x`

`=>a^x = b`

No2.

`log_a b + log_a c = log _a bc`

Now we are good to go

`[log_4(x+3)] + [log_4(2-x)] = 1`

`log_4(x+3) + log_4(2-x) = 1`

Now lets Identify our common base`=> 4`

Now

Using rule number 2 Lets combine

`=>log _4 (x+3)(2-x) = 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 `(x+3)(2-x)`

So lets wrap it up;

`4^1 = (x+3)(2-x)`

`4 = (x+3)(2-x)`

Lets now distribute and multiply;

`(x+3)(2-x) = (2)(x) + (-x)(x) + (3)(2) + (3)(-x)`

`(x+3)(2-x) = 2x -x^2 + 6 -3x`

`(x+3)(2-x) = -x^2 + 6 -x`

=> `-x^2 + 6 -x = 4`

=>`-x^2 - x + 2 = 0`

Now lets try factoring

So

`a*b = -2`, `a+b = -1`

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

=>`-x^2 - 2x + x + 2 = 0`

`-x(x+2) + 1(x + 2) = 0`

`=>(1-x)(x+2) = 0`

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

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Check
=

`=>log _4 (x+3)(2-x) = 1`
......................................................................................................................
x = 1

`=>log _4 (1+3)(2-1) = 1`
`=>log _4 4 = 1`

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

`=>log _4 (-2+3)(2-(-2)) = 1`
`=>log _4 (1)(2+2) = 1`

`=>log _4 4 = 1`

Which is right ; So it is a valid solution

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