The problem asks us to solve the exponential equation $(2x-3)^{-3/2} = 5$ without using logarithms.

1. Problem Description

The problem asks us to solve the exponential equation $(2x-3)^{-3/2} = 5$ without using logarithms.

2. Solution Steps

We are given the equation $(2x-3)^{-3/2} = 5$.

To solve for $x$, we can raise both sides of the equation to the power of $-\frac{2}{3}$.

$((2x-3)^{-3/2})^{-2/3} = 5^{-2/3}$

Using the rule $(a^m)^n = a^{mn}$, we have:

$(2x-3)^{(-3/2)(-2/3)} = 5^{-2/3}$

$(2x-3)^{1} = 5^{-2/3}$

$2x-3 = 5^{-2/3}$

$2x-3 = \frac{1}{5^{2/3}}$

$2x-3 = \frac{1}{\sqrt[3]{5^2}}$

$2x-3 = \frac{1}{\sqrt[3]{25}}$

$2x = 3 + \frac{1}{\sqrt[3]{25}}$

$2x = 3 + \frac{1}{25^{1/3}}$

$x = \frac{1}{2} \left(3 + \frac{1}{25^{1/3}} \right)$

$x = \frac{3}{2} + \frac{1}{2\sqrt[3]{25}}$

$x = \frac{3}{2} + \frac{1}{2 \cdot 25^{1/3}}$

Now we can rationalize the denominator:

$x = \frac{3}{2} + \frac{1}{2 \sqrt[3]{25}} \cdot \frac{\sqrt[3]{5}}{\sqrt[3]{5}}$

$x = \frac{3}{2} + \frac{\sqrt[3]{5}}{2 \sqrt[3]{125}}$

$x = \frac{3}{2} + \frac{\sqrt[3]{5}}{2 \cdot 5}$

$x = \frac{3}{2} + \frac{\sqrt[3]{5}}{10}$

$x = \frac{15}{10} + \frac{\sqrt[3]{5}}{10}$

$x = \frac{15+\sqrt[3]{5}}{10}$

3. Final Answer

$x = \frac{15+\sqrt[3]{5}}{10}$