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

1 Answer
Nov 29, 2015

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

=========================================================
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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