Question

Hey, Indices is a really confusing chapter and there are a few questions I'm unable to solve. this is one of them. can u pls help
thank you!!
- Prarthana​

Answer 1

⭐⭐Given:⭐⭐

$\frac{(\sqrt{32})^{x} }{ {2}^{y + 1} } = 1 \: \: .........(1)$

and

${8}^{y} - 16^{4 - \frac{x}{2} } = 0 \: \: .......(2)$

⭐⭐Answer:⭐⭐

⭐Explanation:⭐

Firstly, we need to write equation 1 in base 2.

We can write,

$\sqrt{32} = {32}^{\frac{1} {2}x }$

$\frac{1}{2}$ is for square root.

$32^\frac{1} {2} x = 2^{(5 \times \frac{1}{2}) x} \implies \: 2^{ \frac{5}{2} x}$

$\implies \: \frac{2 ^{ \frac{5}{2} x} }{2 ^{y + 1} } = 1$

Cross multiplying, we get:

$2^{ \frac{5}{2} x} = 2^{y + 1}$

Comparing the bases , we get:

$y + 1 = \frac{5}{2} x$

Cross multiplying, we get:

$- 5x + 2y = - 2 \: \: .........(3)$

Now , we will write equation (2) in base of 2.

We can write:

$2^{3 \times y} - 16^{4 - \frac{x}{2} } = 0$

$2^{3y} - 16^{ \frac{8 - x}{2} } = 0$

$2^{3y} - 2^{4(\frac{8 - x}{2} )} = 0$

$2^{3y} - 2^{2(8 - x)} = 0$

$2^{3y} - 2^{16 - 2x} = 0$

$2^{3y} = 2^{16 - 2x}$

Comparing bases , we get:

$16 - 2x = 3y$

$2x + 3y = 16 \: \: ..........(4)$

Multiplying (3) by 2 and (4) by 5 , we get:

$- 10x + 4y = - 4$

and

$10x + 15y = 80$

Adding the above two equation:

19y = 76

y = 4

Substituting the value of y in equation (4), we get :

2x+12=16

2x=4

x=2