If $$\frac{8^3. (27)^4. 6^5}{(36)^2.9^4.(18)^2} = 2^{\alpha}.3^{\beta}$$ then $$\alpha + \beta =$$

Solution

$$\frac{8^3. (27)^4. 6^5}{(36)^2.9^4.(18)^2}$$ =$$\frac{\left(\left(2^3\right)^3\ \left(3^3\right)^4\left(2\cdot3\right)^5\right)}{\left(2^2.3^2\right)\left(3^2\right)^4\left(2\cdot3^2\right)}\ =\frac{\left(2^9.3^{12}.2^5.3^5\right)}{\left(2^4.3^4.3^8.2^2.3^4\right)}=\frac{\left(2^{9+5}.3^{12+5}\right)}{\left(2^6.3^{16}\right)}\ =\ 2^{14-6}.3^{17-16}\ =\ 2^8.3^1$$

On comparing,

$$\alpha\ \ =\ 8,\ \beta\ \ =\ 1.\ And\ \alpha\ \ +\ \beta\ \ =\ 9$$

Hence 9 Answer