Sure, let's address each question step-by-step:
### Question 1:
Solve [tex]\(2 \times 4^{1-x} = 8^{-x}\)[/tex].
To solve the equation, we can rewrite both sides of the equation using powers of 2.
1. [tex]\(4\)[/tex] is [tex]\(2^2\)[/tex], so [tex]\(4^{1-x} = (2^2)^{1-x} = 2^{2(1-x)} = 2^{2 - 2x}\)[/tex].
2. [tex]\(8\)[/tex] is [tex]\(2^3\)[/tex], so [tex]\(8^{-x} = (2^3)^{-x} = 2^{3(-x)} = 2^{-3x}\)[/tex].
Substituting these back into the equation:
[tex]\[2 \cdot 2^{2 - 2x} = 2^{-3x}\][/tex]
Combine the terms on the left-hand side:
[tex]\[2^1 \cdot 2^{2 - 2x} = 2^{-3x}\][/tex]
[tex]\[2^{1 + 2 - 2x} = 2^{-3x}\][/tex]
[tex]\[2^{3 - 2x} = 2^{-3x}\][/tex]
Since the bases are the same, we can equate the exponents:
[tex]\[3 - 2x = -3x\][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[3 - 2x = -3x\][/tex]
Add [tex]\(2x\)[/tex] to both sides:
[tex]\[3 = -x\][/tex]
Multiply both sides by [tex]\(-1\)[/tex]:
[tex]\[x = -3\][/tex]
So, the answer to the first question is:
[tex]\[ \boxed{-3} \][/tex]
### Question 2:
Points [tex]\(P, Q\)[/tex], and [tex]\(R\)[/tex] are in the same plane in a field, [tex]\(PQ = 25 \, \text{cm}\)[/tex], [tex]\(PR = 24 \, \text{cm}\)[/tex], and [tex]\(\triangle QRP = 90^{\circ}\)[/tex]. Calculate the length of [tex]\(QR\)[/tex].
Given:
- [tex]\(PQ = 25 \, \text{cm}\)[/tex]
- [tex]\(PR = 24 \, \text{cm}\)[/tex]
- [tex]\(\angle QRP = 90^{\circ}\)[/tex]
We can use the Pythagorean theorem to find [tex]\(QR\)[/tex].
In [tex]\(\triangle QRP\)[/tex] which is a right-angled triangle:
[tex]\[ PQ^2 = PR^2 + QR^2 \][/tex]
Substitute the known values:
[tex]\[ 25^2 = 24^2 + QR^2 \][/tex]
Calculate the squares:
[tex]\[ 625 = 576 + QR^2 \][/tex]
Rearrange to solve for [tex]\(QR^2\)[/tex]:
[tex]\[ QR^2 = 625 - 576 \][/tex]
[tex]\[ QR^2 = 49 \][/tex]
Take the square root of both sides:
[tex]\[ QR = \sqrt{49} \][/tex]
[tex]\[ QR = 7 \, \text{cm} \][/tex]
So, the answer to the second question is:
[tex]\[ \boxed{7 \, \text{cm}} \][/tex]
Both questions have been solved step-by-step as requested.
