Question

See screenshot below​

Answer 1

A&D

Explanation:

we want to solve the following equation for x:

`\displaystyle ( {2}^(x) - 3) ({2}^(x) - 4) = 0`

to do so let
`2^x` be u and transform the equation:

`\displaystyle (u - 3) (u - 4) = 0`

By Zero product property we obtain:

`\displaystyle \begin{cases}u - 3 = 0\\ u - 4= 0 \end{cases}`

Solve the equation for u which yields:

`\displaystyle \begin{cases}u = 3\\ u = 4 \end{cases}`

substitute back:

`\displaystyle \begin{cases} {2}^(x) = 3\\ {2}^(x) = 4 \end{cases}`

take logarithm of Base 2 in both sides of the both equations:

`\displaystyle \begin{cases} \log_(2) {2}^(x) = \log_(2) 3\\ \log_(2) {2}^(x) = \log_(2) 4 \end{cases}`

hence,

`\displaystyle \begin{cases} x_(1) = \log_(2) 3\\ x_(2) = 2 \end{cases}`

Answer 2

( A ) and ( D )

Explanation:

( 2^x -3 ) (
`2^x -4` ) = 0

• When the the product of factors equals 0, atleast one factor is 0.

`2^x`- 3 = 0

2^x - 4 = 0

• Solve for x.

`2^x`- 3 = 0 ,
`2^x -4`

x =
`log_2` (3) , x = 2

The equation has two solutions;

`x_1` =
`log_2` (3) ,
`x_2` = 2.