1. **State the problem:** Solve the given exponential and logarithmic expressions and simplify the fraction involving powers of 5 and 25. 2. Evaluate the expression $x \log 6 = 7 \times 13$. We have $x \log 6 = 91$ because $7 \times 13 = 91$. To isolate $x$, divide both sides by $\log 6$: $$x = \frac{91}{\log 6}$$ 3. Solve for $x$ in the equation $6^x = 13$. Take the logarithm (base 10 or natural log) of both sides: $$x \log 6 = \log 13$$ Hence, $$x = \frac{\log 13}{\log 6}$$ 4. Simplify the expression: $$\frac{5^{x-y} \times 125^{3x-y}}{25^x}$$ First, express all bases as powers of 5: $$125 = 5^3, \quad 25 = 5^2$$ Substitute: $$\frac{5^{x-y} \times (5^3)^{3x-y}}{(5^2)^x} = \frac{5^{x-y} \times 5^{3(3x-y)}}{5^{2x}}$$ Simplify the exponents: $$5^{x-y} \times 5^{9x - 3y} = 5^{(x-y) + (9x - 3y)} = 5^{10x - 4y}$$ The denominator is: $$5^{2x}$$ So the entire expression is: $$\frac{5^{10x - 4y}}{5^{2x}} = 5^{(10x - 4y) - 2x} = 5^{8x - 4y}$$ 5. **Final answers:** - $x$ from $x \log 6 = 91$ is $$x = \frac{91}{\log 6}$$ - $x$ from $6^x = 13$ is $$x = \frac{\log 13}{\log 6}$$ - Simplified fraction is $$5^{8x - 4y}$$