Question

please solve it fast I need irgent​

Answer 1

Question:

Solve the following equations for x:

(i) $2^{2x} - 2^{x+3} + 2^4 = 0$

This is very simple. Don't be confused with the exponent.

Here, we can simplify it for computing the value of x.

$\implies 2^{2x} - 2^{x+3} + 2^4 = 0$

We can write it as follows because when we'll multiply it, it will be in its original form.

$\implies (2^x)^{2} - (2^x)(2^3) + 16 = 0$

$\implies (2^x)^{2} - (2^x)(8) + 16 = 0$

As we can see that, 2ˣ and 2ˣ are same in both the terms just for making it in the form of variables so that, we can factorise it.

So, let 2ˣ = y

$\implies y^2 - 8y + 16 = 0$

By using splitting the middle term method:

Step 1: For factorsing any polynomial, just write it as it is. If it's not in its general form try to write it in its general form without changing its sign.

y² - 8y + 16

Step 2: Multiply the co-efficient of y² and the constant term.

Here, the co-efficient of y² is 1 and the constant term is 16.

Multiplying 1 and 16 we obtain 16.

Step 3: Now, we've to find the two factors of 16 which would give us the sum as the co-efficient of y.

Here, the co-efficient of y is - 8.

Step 4: Find the factors of 16 which would gives us the sum as 8.

The factors of 16 are $\pm 1, \pm 2, \pm 4, \pm 8 \pm 16$

So, the two factors of 16 which would give us the sum as - 8 are - 4 and - 4.

We can also repeat the numbers.

Step 5: Substitute the factors in the place of - 8y along with the respective variables.

$\implies y^2 - 8y + 16$

$\implies y^2 - 4y - 4y + 16$

Step 6: Group the terms.

$\implies (y^2 - 4y) - (4y + 16)$

Step 7: Take out the common terms.

$\implies y(y - 4) - 4(y - 4)$

Step 8: You'll observe that, both the terms inside the terms will be same as the other. Write the common terms first and then write the uncommon terms.

$\implies (y-4)(y-4)$

Therefore, (y - 4) and (y - 4) are the factors of y² - 8y + 16.

As given, y² - 8y + 16 = 0, therefore, it's factors must be equal to 0.

$\therefore \implies y - 4 = 0 \:\&\: y - 4 = 0$

$\boxed{\bold{\pink{\mathsf{\implies y = 4}}}}$

We actually don't need to find the value of y from the second equation i.e., y - 4 = 0, as we gotta the value from first equation. As the variable is same, there values must be same.

Resubstituting y = 2ˣ

2ˣ = y

2ˣ = 4

2ˣ = 2²

$(\cancel{2})^x = (\cancel{2})^2$

$\boxed{\bold{\red{\mathsf{x = 2}}}}$

Therefore, the value of x is 2.

Answer 2

HEY USER HERE IS YOUR ANSWER IN THE ATTACHMENT GIVEN ABOVE