Surds, Indices and Logarithms

$75 = 25 \times 3$ and 25 is a perfect square, so $\sqrt{75} = \sqrt{25}\sqrt3 = 5\sqrt3$.

Ask what power of 3 gives 81. Since $3^4 = 81$, the logarithm is 4.

The numerator is $8x^9y^{-6}$. Dividing: $8 \div 4 = 2$, $x^{9-2} = x^7$, and $y^{-6-1} = y^{-7}$. So the answer is $2x^7y^{-7} = \dfrac{2x^7}{y^7}$.

Multiply top and bottom by the conjugate $3+\sqrt5$. The denominator becomes $9 - 5 = 4$, and the numerator becomes $4(3+\sqrt5)$, so the 4s cancel.

Let $y = 3^x$; then $3^{2x} = y^2$ and the equation becomes $y^2 - 10y + 9 = 0$, so $(y-1)(y-9)=0$. Then $3^x = 1$ gives $x = 0$, and $3^x = 9$ gives $x = 2$.

Combining: $\log_5\dfrac{2x+3}{x-1} = 1$, so $\dfrac{2x+3}{x-1} = 5$. Then $2x+3 = 5x-5$, giving $3x = 8$ and $x = \tfrac83$. Check: $x - 1 = \tfrac53 > 0$ and $2x+3 = \tfrac{25}{3}$, and $\tfrac{25}{3} \div \tfrac53 = 5$.