1. **State the problem:** We are given the equation $$(ar^b)^4 = 81^{24}$$ where $a$ and $b$ are positive integers. We need to find the values of $a$ and $b$. 2. **Rewrite the equation:** Using the power of a power rule, $$(ar^b)^4 = a^4 r^{4b}$$ so the equation becomes $$a^4 r^{4b} = 81^{24}$$. 3. **Express 81 as a power of a prime:** Note that $$81 = 3^4$$, so $$81^{24} = (3^4)^{24} = 3^{96}$$. 4. **Rewrite the equation with prime bases:** We want $$a^4 r^{4b} = 3^{96}$$. 5. **Assuming $a$ and $r$ are powers of 3:** Let $$a = 3^x$$ and $$r = 3^y$$ where $x,y$ are positive integers. 6. **Substitute into the equation:** $$a^4 r^{4b} = (3^x)^4 (3^y)^{4b} = 3^{4x} 3^{4by} = 3^{4x + 4by}$$. 7. **Equate exponents:** Since $$3^{4x + 4by} = 3^{96}$$, we have $$4x + 4by = 96$$. 8. **Simplify:** Divide both sides by 4: $$x + by = 24$$. 9. **Find integer solutions:** We want positive integers $x,y,b$ such that $$x + b y = 24$$. 10. **Choose simplest values:** To find $a$ and $b$, we can set $y=1$ (so $r=3$), then $$x + b = 24$$. 11. **Express $a$ and $b$:** Since $a = 3^x$ and $r = 3^1 = 3$, the original expression is $$(a r^b)^4 = (3^x imes 3^b)^4 = 3^{4(x+b)} = 3^{96}$$ which matches if $$x + b = 24$$. 12. **Pick $a$ and $b$:** For simplicity, let $a = 3^{12}$ and $b = 12$ so that $$12 + 12 = 24$$. 13. **Final answer:** $$a = 3^{12} = 531441$$ $$b = 12$$ Thus, $a = 531441$ and $b = 12$ satisfy the equation.