Solve the following equations for x: (ग) 2^x+1 + 2^x+2 + 2^x+3 = 448 (ड) 4^x+1 - 8 x 4^x−1 = 32 (ढ) 3^x+2 + 3^x+1 + 2 x 3^x = 126 (फ) 8^x−1 − 23x−2 + 8 = 0

$$2^x \cdot 2^1 + 2^x \cdot 2^2 + 2^x \cdot 2^3 = 448$$ $$2 \cdot 2^x + 4 \cdot 2^x + 8 \cdot 2^x = 448$$

$$2^x (2 + 4 + 8) = 448$$ $$2^x (14) = 448$$

$$2^x = \frac{448}{14}$$ $$2^x = 32$$

$$x = 5$$

The solution for (ग) is $\boxed{x=5}$.

Equation (ड):

$$4^{x+1} - 8 \times 4^{x-1} = 32$$

$$4^x \cdot 4^1 - 8 \cdot 4^x \cdot 4^{-1} = 32$$ $$4 \cdot 4^x - 8 \cdot 4^x \cdot \frac{1}{4} = 32$$ $$4 \cdot 4^x - 2 \cdot 4^x = 32$$

$$4^x (4 - 2) = 32$$ $$4^x (2) = 32$$

$$4^x = \frac{32}{2}$$ $$4^x = 16$$

$$x = 2$$

The solution for (ड) is $\boxed{x=2}$.

Equation (छ):

$$3^{x+2} + 3^{x+1} + 2 \times 3^x = 126$$

$$3^x \cdot 3^2 + 3^x \cdot 3^1 + 2 \cdot 3^x = 126$$ $$9 \cdot 3^x + 3 \cdot 3^x + 2 \cdot 3^x = 126$$

$$3^x (9 + 3 + 2) = 126$$ $$3^x (14) = 126$$

$$3^x = \frac{126}{14}$$ $$3^x = 9$$

$$x = 2$$

The solution for (छ) is $\boxed{x=2}$.

Equation (फ):
The equation is $8^{x-1} - 23^{x-2} + 8 = 0$. Given the context of the other problems, it is highly probable that the term $23^{x-2}$ is actually $23 \times 8^{x-2}$. We will proceed with this assumption.

Assume the equation is:

$$8^{x-1} - 23 \times 8^{x-2} + 8 = 0$$

$$8^x \cdot 8^{-1} - 23 \cdot 8^x \cdot 8^{-2} + 8 = 0$$ $$\frac{8^x}{8} - 23 \cdot \frac{8^x}{64} + 8 = 0$$

Substitute $y$ into the equation.

$$\frac{y}{8} - \frac{23y}{64} + 8 = 0$$

$$64 \left( \frac{y}{8} \right) - 64 \left( \frac{23y}{64} \right) + 64(8) = 0$$ $$8y - 23y + 512 = 0$$

$$-15y = -512$$ $$y = \frac{512}{15}$$

$$x = \log_8 \left( \frac{512}{15} \right)$$

Since $512 = 8^3$, we can write:

$$x = \log_8 (8^3) - \log_8 (15)$$ $$x = 3 - \log_8 (15)$$

The solution for (फ) is $\boxed{x = 3 - \log_8 (15)}$.