Solve the following exponential equations for x: a) 27^x = 93x−2 b) 32^x+1 = 42x−1 c) 81^x−2 = 93x+1. Also solve: a) 32x+1 - 9^x = 162

$$3^{3x} = 3^{2(3x-2)}$$ $$3^{3x} = 3^{6x-4}$$

$$4 = 6x - 3x$$ $$4 = 3x$$ $$x = \frac{4}{3}$$

The solution is $\boxed{x = \frac{4}{3}}$.

b) $32^{x+1} = 4^{2x-1}$

$$2^{5(x+1)} = 2^{2(2x-1)}$$ $$2^{5x+5} = 2^{4x-2}$$

$$5x - 4x = -2 - 5$$ $$x = -7$$

The solution is $\boxed{x = -7}$.

c) $81^{x-2} = 9^{3x+1}$

$$9^{2(x-2)} = 9^{3x+1}$$ $$9^{2x-4} = 9^{3x+1}$$

$$-4 - 1 = 3x - 2x$$ $$-5 = x$$

The solution is $\boxed{x = -5}$.

3. Solve for $x$:

a) $3^{2x+1} - 9^x = 162$

Note that $9 = 3^2$.

$$3^{2x} \cdot 3^1 - (3^2)^x = 162$$ $$3 \cdot (3^x)^2 - (3^x)^2 = 162$$

Substitute $y$ into the equation.

$$3y^2 - y^2 = 162$$ $$2y^2 = 162$$

$$y^2 = \frac{162}{2}$$ $$y^2 = 81$$ $$y = \pm\sqrt{81}$$ $$y = \pm 9$$

Since $3^x$ must be positive, we take $y=9$.

$$3^x = 9$$ $$3^x = 3^2$$ $$x = 2$$

The solution is $\boxed{x = 2}$.

b) $8^{x+1} - 2^{3x-1} = 120$

Note that $8 = 2^3$.

$$(2^3)^{x+1} - 2^{3x-1} = 120$$ $$2^{3(x+1)} - 2^{3x-1} = 120$$ $$2^{3x+3} - 2^{3x-1} = 120$$

$$2^{3x} \cdot 2^3 - 2^{3x} \cdot 2^{-1} = 120$$ $$2^{3x} (2^3 - 2^{-1}) = 120$$ $$2^{3x} \left( 8 - \frac{1}{2} \right) = 120$$

$$2^{3x} \left( \frac{16}{2} - \frac{1}{2} \right) = 120$$ $$2^{3x} \left( \frac{15}{2} \right) = 120$$

$$2^{3x} = 120 \cdot \frac{2}{15}$$ $$2^{3x} = \frac{240}{15}$$ $$2^{3x} = 16$$

$$2^{3x} = 2^4$$ $$3x = 4$$ $$x = \frac{4}{3}$$

The solution is $\boxed{x = \frac{4}{3}}$.

c) $7^{2x+1} - 49^x = 294$

Note that $49 = 7^2$.

$$7^{2x} \cdot 7^1 - (7^2)^x = 294$$ $$7 \cdot (7^x)^2 - (7^x)^2 = 294$$

Substitute $y$ into the equation.

$$7y^2 - y^2 = 294$$ $$6y^2 = 294$$

$$y^2 = \frac{294}{6}$$ $$y^2 = 49$$ $$y = \pm\sqrt{49}$$ $$y = \pm 7$$

Since $7^x$ must be positive, we take $y=7$.

$$7^x = 7$$ $$7^x = 7^1$$ $$x = 1$$

The solution is $\boxed{x = 1}$.