Hello, Danielisew!

For most of these, you're expected to know that:

. . If $\displaystyle b^x\:=\:b^y$, then $\displaystyle x = y.$

Given two equal exponentials with equal bases, their exponents are equal.

Quote:

1a) Given that $\displaystyle 2^a \,= \,8^b$, express $\displaystyle a$ in terms of $\displaystyle b.$

We have: .$\displaystyle 2^a\:=\:(2^3)^b\quad\Rightarrow\quad 2^a\:=\:2^{3b}$

Therefore: .$\displaystyle \boxed{a \:=\:3b}$

Quote:

1b) State the value of $\displaystyle x$ for which $\displaystyle 9^x \,= \,3.$

We have: .$\displaystyle (3^2)^x\:=\:3\quad\Rightarrow\quad3^{2x} \:=\:3^1$

Therefore: .$\displaystyle 2x \:=\:1\quad\Rightarrow\quad\boxed{ x \:=\:\frac{1}{2}}$

Quote:

2) If $\displaystyle x^p = (\sqrt{x})^3,\;x^q = \frac{1}{x^2}\;\text{and }x^r = \frac{x^p}{x^q}$

a) Find the value of $\displaystyle x$ . . . . Did you mean $\displaystyle {\color{blue}r}$ ?

$\displaystyle x^p \:=\:(\sqrt{x})^3\quad\Rightarrow\quad x^p \:=\:x^{\frac{3}{2}}\quad\Rightarrow\quad p \:=\:\frac{3}{2}$

$\displaystyle x^q \:=\:\frac{1}{x^2}\quad\Rightarrow\quad x^q \:=\:x^{-2}\quad\Rightarrow\quad q \:=\:-2$

Then: .$\displaystyle x^r\:=\:\frac{x^{\frac{3}{2}}}{x^{-2}}\quad\Rightarrow\quad x^r\:=\:x^{\frac{7}{2}}\quad\Rightarrow\quad\boxed { r \:=\:\frac{7}{2}}$

Quote:

b) Evaluate $\displaystyle x^r$ when $\displaystyle x = 4$

$\displaystyle \text{If }x = 4\!:\;\;x^r \;=\;4^{\frac{7}{2}} \;=\;\boxed{128}$