(2, 14), (6, 34), (8,44), (x, 64). What is the value of x?

$\begin{gathered} 2\log _3x-\log _3(x-2)=2 \\ \log _3x^2-\log _3(x-2)=2_{}\text{ \lbrack{}log(a)-log(b) = log(a/b)\rbrack} \\ \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \end{gathered}$

Simplify further.

$\begin{gathered} \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \\ \frac{x^2}{x-2}=3^2 \\ x^2=9(x-2) \\ x^2-9x+18=0 \end{gathered}$

Solve the quadratic equation for x.

$\begin{gathered} x^2-6x-3x+18=0 \\ x(x-6)-3(x-6)=0 \\ (x-6)(x-3)=0 \end{gathered}$

From the above equation (x - 6) = 0 or (x - 3) = 0.

For (x - 6) = 0,

$\begin{gathered} x-6=0 \\ x=6 \end{gathered}$

For (x - 3) = 0,

$\begin{gathered} x-3=0 \\ x=3 \end{gathered}$

The values of x from solving the equations are x = 3 and x = 6.

Substitute the values of x in the equation to check answers are valid or not.

For x = 3,

$\begin{gathered} 2\log _3(3^{})-\log _3(3-2)=2 \\ 2\log _33-\log _31=2 \\ 2\cdot1-0=2 \\ 2=2 \end{gathered}$

Equation satisfy for x = 3. So x = 3 is valid value of x.

For x = 6,

$\begin{gathered} 2\log _36-\log _3(6-2)=2 \\ 2\log _36-\log _34=2 \\ \log _3(6^2)-\log _34=2 \\ \log _3(\frac{36}{4})=2 \\ \log _39=2 \\ \log _3(3^2)=2 \\ 2\log _33=2 \\ 2=2 \end{gathered}$

Equation satifies for x = 6.

Thus values of x for equation are x = 3 and x = 6.

6 0

1 year ago

It is a well-known fact that 50% of the general population are rascals (P(R) = 0.5) and 50% of the general population are not ra

aleksley [76]

Answer:

The answer is D. 0.75

Step-by-step explanation:

Let call R the event that a person is rascal, RC a person is not rascal, TH a person use top hat and NTH a person don’t use top hat.

From the information on the question we have 4 options with their respective probability:

1. A person could be rascal and use top Hat: this probability is calculate as the multiplication of the probability of be a rascal (0.5) and use a top hat (0.9), then:

P(R y TH)=0.5*0.9=0.45

2. A person could be rascal and don’t use top Hat: this probability is calculate as the multiplication of the probability of be a rascal (0.5) and not use a top hat (0.9), then:

P(R y NTH)=0.5*0.1=0.05

3. A person could be not rascal and use top Hat: this probability is calculate as the multiplication of the probability of not be a rascal (0.5) and use a top hat (0.3), then:

P(RC y TH)=0.5*0.3=0.15

4. A person could be not rascal and not use top Hat: this probability is calculate as the multiplication of the probability of not be a rascal (0.5) and not use a top hat (0.7), then:

P(RC y NTH)=0.5*0.7=0.35

Then the probability that a person is a rascal given that he is wearing a top hat could be written and calculate as:

$P(R/TH)=\frac{P(R y TH)}{P(TH)}$

For calculate P(TH) we need to sum all the option in which TH is involve so:

P(TH) = P(R y TH)+ P(RC y TH)

P(TH)=0.45+0.15=0.6

Replacing values on the first equation we get:

$P(R/TH)=\frac{0.45}{0.6} =0.75$

So, the probability that a person is a rascal given that he is wearing a top hat is 0.75

5 0

3 years ago