At first, we will break the parenthesis. To break that, we will multiply the value inside the parenthesis by the adjacent number, that is 5. Again, we have to consider the Algebraic operation (Golden rule) -

[(-) x (-) = (+); (+) x (-) = (-)]

Therefore, since there is a minus sign in each of the value inside the parenthesis, the result will be minus as 5 is a positive integer.

or, -5*(2k) - (5*3) + 2k

or, -10k - 15 + 2k

or, -8k - 15 (after the deduction)

The answer is = -8k - 15

6 0

2 years ago

hi, i dont undertand number 20 because i was absent in class today and i rerally need help, i will appraciate with the help, and

Mariulka [41]

Given:

The equation is,

$2\log _3x-\log _3(x-2)=2$

Explanation:

Simplify the equation by using logarthimic property.

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

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