Question

solve these all
please don't let my points to go in vain

Answer 1

Actually I can't find the solution to the most, but I'm writing only as a pathway to find the right. Hope you may get the answer through this. 

1. The answer is 3 (Proved earlier). And 0 is also the answer. I got this by the option! 

3. $\frac{2^x}{1 + 2^x} = \frac{1}{4} \\ \\ 2^x * 4 = 2^x + 1$ 

Let $2^x$ be n. 

$4n = n + 1 \\ 4n - n = 1 \\ = 3n = 1 \\ \\ n = \frac{1}{3}$ 

∴ $2^x = \frac{1}{3}$ 

$8^x = (2^3)^x = (2^x)^3 = (\frac{1}{3})^3 = \frac{1}{27} \\ \\ 1 + 8^x = 1 + \frac{1}{27} = \frac{28}{27} \\ \\ \frac{8^x}{1 + 8^x} = \frac{1}{27} / \frac{28}{27} = \frac{1}{27} * \frac{27}{28} = \frac{1}{28}$ 

4. If $a^{\frac{1}{x}} = b^{\frac{1}{y}} = c^{\frac{1}{2}} = k$, 

$a = k^x \\ b = k^y \\ c = k^2$ 

$k^x * k^2 = k^{x + 2} = ac \\ \\ (k^y)^2 = k^{2y} = b^2 \\ \\ k^{x + 2} * k^{2y} = k^{x + 2 + 2y} = b^2 ac \\ \\ (k^{x + 2 + 2y})^{\frac{1}{2y}} = k^{\frac{x + 2}{2y} + 1} = (b^2 ac)^{\frac{1}{2y}} \\ \\ \frac{k^{\frac{x + 2}{2y} + 1}}{k} = k^{\frac{x + 2}{2y}} = \frac{(b^2 ac)^{\frac{1}{2y}}}{k}$ 

I'd done as much as I can. Hope the answers can be found through these.