Evaluate the following: 32^(3/2), 25^(1/5), 216^(4/3), 19°

Substitute this into the equation.

$$y^2 + 2y - 3 = 0$$

We can factor the quadratic expression.

$$(y+3)(y-1) = 0$$

This gives two possible values for $y$:

$$y+3 = 0 \implies y = -3$$ $$y-1 = 0 \implies y = 1$$

For $y = -3$:

$$3^x = -3$$

An exponential function $3^x$ can never be negative, so there is no real solution for $x$ in this case.

For $y = 1$:

$$3^x = 1$$

Since any non-zero number raised to the power of 0 is 1, we have:

$$3^x = 3^0$$

Therefore,

$$x = 0$$

The solution for the equation is:

$$\boxed{x = 0}$$

$$32^{3/2} = (\sqrt{32})^3 = (4\sqrt{2})^3 = 4^3 \cdot (\sqrt{2})^3 = 64 \cdot (2\sqrt{2}) = 128\sqrt{2}$$ $$\boxed{128\sqrt{2}}$$

$$25^{1/5} = (5^2)^{1/5} = 5^{2/5}$$ $$\boxed{5^{2/5}}$$

$$216^{4/3} = (\sqrt[3]{216})^4$$

Since $6^3 = 216$, we have $\sqrt[3]{216} = 6$.

$$(\sqrt[3]{216})^4 = 6^4 = 1296$$ $$\boxed{1296}$$

Any non-zero number raised to the power of 0 is 1.

$$19^0 = 1$$ $$\boxed{1}$$