Question

What is the solution to 2logg(x) = logg8+log,(x-2)? x=-4 x=-2 X=4 X= 8

Answer 1

Answer:

x = 4

which agrees with the third listed answer among the given options

Step-by-step explanation:

We can re-write the logarithmic equation using the properties of logarithms:

$2\,log(x)=log(8)+log(x-2)\\log(x^2)=log(8*(x-2))\\log(x^2)=log(8x-16)$

Therefore, the arguments of the log functions must also be equal (and we can solve for "x" by noticing that this expression is the perfect square of a binomial:

$x^2=8x-16\\x^2-8x+16=0\\(x-4)^2=0$

and for this equation to verify, x must be 4

Answer 2

Answer:

The value of x is 4.

Step-by-step explanation:

You have to use Logarithm Laws :

$log( {a}^{n} ) \: ⇒ \: n log(a)$

$log(a) + log(b) \: ⇒ \: log(a \times b)$

So for this question :

$2 log(x) \: ⇒ \: log( {x}^{2} )$

$log(8) + log(x - 2) \: ⇒ \: log(8(x - 2))$

$log( {x}^{2} ) = log(8x - 16)$

Next you have to cut out the log :

${x}^{2} = 8x - 16$

Then, you have to make the equation equals 0 :

${x}^{2} - 8x + 16 = 0$

Lastly, you have to solve it :

${x}^{2} - 4x - 4x + 16 = 0$

$x(x - 4) - 4(x - 4) = 0$

$(x - 4)(x - 4) = 0$

$x = 4$