Here are the solutions to your problems. b) Given the function $f(t) = 1 - \frac{3}{t}$ and the range $\{-1, 3, 7\}$, we need to find the domain. The range consists of the output values of $f(t)$. We set $f(t)$ equal to each value in the range and solve for $t$. Step 1: Set $f(t) = y$ and solve for $t$ in terms of $y$. $$y = 1 - \frac{3}{t}$$ $$y - 1 = -\frac{3}{t}$$ $$\frac{3}{t} = 1 - y$$ $$t = \frac{3}{1 - y}$$ Step 2: Substitute each value from the range into the expression for $t$. For $y = -1$: $$t = \frac{3}{1 - (-1)} = \frac{3}{1 + 1} = \frac{3}{2}$$ For $y = 3$: $$t = \frac{3}{1 - 3} = \frac{3}{-2} = -\frac{3}{2}$$ For $y = 7$: $$t = \frac{3}{1 - 7} = \frac{3}{-6} = -\frac{1}{2}$$ The domain is the set of these $t$ values. The domain is $\boxed{\left\{-\frac{3}{2}, -\frac{1}{2}, \frac{3}{2}\right\}}$. a) Given $a^{x-3}b^{3x} = a^{5-x}b^{5x}$, we need to show that $x \log\left(\frac{a}{b}\right) = 4 \log a$. Step 1: Group terms with the same base on opposite sides of the equation. $$\frac{a^{x-3}}{a^{5-x}} = \frac{b^{5x}}{b^{3x}}$$ Step 2: Apply the exponent rule $\frac{m^p}{m^q} = m^{p-q}$. $$a^{(x-3) - (5-x)} = b^{5x - 3x}$$ $$a^{x-3-5+x} = b^{2x}$$ $$a^{2x-8} = b^{2x}$$ Step 3: Take the logarithm of both sides. $$\log(a^{2x-8}) = \log(b^{2x})$$ Step 4: Apply the logarithm rule $\log(M^p) = p \log M$. $$(2x-8)\log a = 2x \log b$$ Step 5: Distribute and rearrange the terms to isolate $x$. $$2x \log a - 8 \log a = 2x \log b$$ $$2x \log a - 2x \log b = 8 \log a$$ Step 6: Factor out $2x$ from the left side. $$2x (\log a - \log b) = 8 \log a$$ Step 7: Apply the logarithm rule $\log M - \log N = \log\left(\frac{M}{N}\right)$. $$2x \log\left(\frac{a}{b}\right) = 8 \log a$$ Step 8: Divide both sides by 2. $$x \log\left(\frac{a}{b}\right) = 4 \log a$$ This completes the proof.