Respuesta :

highwireact

Answer:

B. x = 4

Step-by-step explanation:

I can't speak to the first part of this question, as I don't totally have context for what they're asking, but we can solve for x using one of the laws of logarithms, namely:

[tex]\log_bm+\log_bn=\log_bmn[/tex]

Using this law, we can combine and rewrite our initial equation as

[tex]\log_2(x\cdot(x-2))=3\\\log_2(x^2-2x)=3[/tex]

Remember that logarithms are simply another way of writing exponents. The logarithm [tex]\log_28=3[/tex] is just another way of writing the fact [tex]2^3=8[/tex]. Keeping that in mind, we can express our logarithm in terms of exponents as

[tex]\log_2(x^2-2x)=3\rightarrow2^3=x^2-2x[/tex]

2³ = 8, so we can replace the left side of our equation with 8 to get

[tex]8 = x^2-2x[/tex]

Moving the 8 to the other side:

[tex]0=x^2-2x-8[/tex]

We can now factor the expression on the right to find solutions for x:

[tex]0=(x-4)(x+2)\\x=4, -2[/tex]

The only option which agrees with our solution is B.

Answer:

The answer is:

        [tex]y_1=\dfrac{\log x}{\log 2}+\dfrac{\log (x-2)}{\log 2}\ ,\ y_2=3\ ,\ x=4[/tex]

Step-by-step explanation:

We are given a logarithmic expression as:

[tex]\log_2 x+log_2 (x-2)=3[/tex]

As we know that:

[tex]\log_a x=\dfrac{\log x}{\log a}[/tex]

Hence, we get the logarithmic expression as follows:

[tex]\dfrac{\log x}{\log 2}+\dfrac{\log (x-2)}{\log 2}=3[/tex]

We know that we can get the system of equations as follows:

[tex]y_1=\dfrac{\log x}{\log 2}+\dfrac{\log (x-2)}{\log 2}[/tex]

and

[tex]y_2=3[/tex]

Hence, when we plot the graph for this system of equations we see that the point of intersection of the graph is: (4,3)

Hence, the solution is the x-value of the point of intersection of the two equations.

Hence, x=4 is the solution.

Hence, the correct option is:

  [tex]y_1=\dfrac{\log x}{\log 2}+\dfrac{\log (x-2)}{\log 2}\ ,\ y_2=3\ ,\ x=4[/tex]

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