Question

Jerome solved the equation below by graphing.

Answer 1

Jerome rewrote the logarithms as

$\log_2x=\dfrac{\log x}{\log2}$

$\log_2(x-2)]=\dfrac{\log(x-2)}{\log 2}$

which eliminates C and D.

Solve the equation:

$\log_2x+\log_2(x-2)=\log_2x(x-2)=3\implies 2^{\log_2x(x-2)}=2^3\implies x(x-2)=8$

$\implies x^2-2x-8=(x-4)(x+2)=0\implies x=4\text{ or }x=-2$

$\log_b(-2)$ is undefined for any base $b$ (where the logarithm is real-valued), so we omit that solution.

This makes B the answer.

Answer 2

Answer:

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

Step-by-step explanation:

Given : Jerome solved the equation below by graphing.

$\log_2x+\log_2(x-2)=3$

To find : Which of the following shows the correct system of equations and solution?

Solution :

Two system of equations formed from given equation,

$y_1=\log_2x+\log_2(x-2)$  

$y_1=\frac{\log x}{\log 2}+\frac{\log (x-2)}{\log 2}$   ......[1]

$y_2=3$  .........[2]

Now, we plot these two equations and the intersection of these two equation is the solution.

The intersection point is (4,3)

Therefore, The solution of given equation is x=3

Hence, Option b is correct.