Question

Please help me with question 47
Give detailed solution
Best answer will be marked brainliest

Answer 1

47.

$Given : 4^x - 3(2^{x + 2}) + 32 = 0$

$\Rightarrow (2^{x})^2 - 3 * 2^x * 2^2 + 32 = 0$

Here, 2ˣ is common anywhere. Let 2ˣ = a.

⇒ a² - 3 * a * 4 + 32 = 0

⇒ a² - 12a + 32 = 0

⇒ a² - 8a - 4a + 32 = 0

⇒ a(a - 8) + 4(a - 8) = 0

⇒ a = 8,4

Now,

(i)

2ˣ = 8

2ˣ = 2³

x = 3

(ii)

2ˣ = 4

2ˣ = 2²

x = 2

Thus, the roots are 2 and 3.

But we have to find the sum of the possible values.

Therefore, there sum is 5.

Hope it helps!

Answer 2

Answer

5

Step-by-step explanation:

$( {2}^{2} ) {}^{x} - 3( {2}^{x} . {2}^{2} ) + 32 = 0 \\( {2}^{x} ) {}^{2} - 12({2}^{x} ) + 32 = 0$

$let \: \: \: \: \: {2}^{x} = y$

Your new equation

${y}^{2} - 12y + 32 = 0 \\ {y }^{2} - 8y - 4y + 32 = 0 \\ (y - 4)(y - 8) = 0 \\ y = 4 \: \: \: and \: \: \: y = 8 \\ {2}^{x} = 4 = {2}^{2} \: \: and \: \: {2}^{x} = 8 = {2}^{3} \\ x = 2 \: \: and \: 3$

Sum = 2 + 3 =5