How do you solve $log_ 4 (x – 6) + log _ 4 x = 2$ ?

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Answer 1 · 2016-01-13T04:56:16

Step by step explanation is given below.

$log_4(x-6)+log_4x=2$

Step 1: Condense the left-hand side using rule
$log_b(P) + log_b(Q) = log_b(PQ)$

Step 2: Convert logarithms to exponent form using the rule
$log_b(a) = k -> a=b^k$

Now let us solve our problem

$log_4(x-6)+log_4(x)=2$

Step 1: Condensing...

$log_4(x-6)x=2$
$log_4(x^2-6x)=2$

Step 2: Converting to exponent form.

$x^2-6x = 2^4$

$=>x^2-6x = 16$

$=>x^2-6x-16=0$

$=>(x-8)(x+2)=0$

$x-8=0$ or $x+2=0$

$x=8 or x = -2$

$x=-2$ is not possible as $log_4(-2)$ is not defined.

The solution is $x=8$

How do you solve log_ 4 (x – 6) + log _ 4 x = 2log_ 4 (x – 6) + log _ 4 x = 2?