Question

Jerome solved the equation below by graphing.

Answer 1

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:

$\log_bm+\log_bn=\log_bmn$

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

$\log_2(x\cdot(x-2))=3\\\log_2(x^2-2x)=3$

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

$\log_2(x^2-2x)=3\rightarrow2^3=x^2-2x$

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

$8 = x^2-2x$

Moving the 8 to the other side:

$0=x^2-2x-8$

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

$0=(x-4)(x+2)\\x=4, -2$

The only option which agrees with our solution is B.

Answer 2

Answer:

The answer is:

        $y_1=\dfrac{\log x}{\log 2}+\dfrac{\log (x-2)}{\log 2}\ ,\ y_2=3\ ,\ x=4$

Step-by-step explanation:

We are given a logarithmic expression as:

$\log_2 x+log_2 (x-2)=3$

As we know that:

$\log_a x=\dfrac{\log x}{\log a}$

Hence, we get the logarithmic expression as follows:

$\dfrac{\log x}{\log 2}+\dfrac{\log (x-2)}{\log 2}=3$

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

$y_1=\dfrac{\log x}{\log 2}+\dfrac{\log (x-2)}{\log 2}$

and

$y_2=3$

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:

  $y_1=\dfrac{\log x}{\log 2}+\dfrac{\log (x-2)}{\log 2}\ ,\ y_2=3\ ,\ x=4$