Question:

Step 1: Convert the logarithmic equations to exponential form. From the definition of logarithms, $\log_b a = c \iff b^c = a$. $$ \log_\gamma(2\alpha) = \frac{1}{3} \implies 2\alpha = \gamma^{1/3} \quad (1) $$ $$ \log_\gamma(5\beta) = \frac{1}{6} \implies 5\beta = \gamma^{1/6} \quad (2) $$ $$ \log_\gamma(\alpha\beta) = \frac{3}{2} \implies \alpha\beta = \gamma^{3/2} \quad (3) $$ Step 2: Express $\alpha$ and $\beta$ in terms of $\gamma$. From equation (1), $\alpha = \frac{\gamma^{1/3}}{2}$. From equation (2), $\beta = \frac{\gamma^{1/6}}{5}$. Step 3: Substitute the expressions for $\alpha$ and $\beta$ into equation (3) to solve for $\gamma$. $$ \left(\frac{\gamma^{1/3}}{2}\right) \left(\frac{\gamma^{1/6}}{5}\right) = \gamma^{3/2} $$ $$ \frac{\gamma^{1/3 + 1/6}}{10} = \gamma^{3/2} $$ $$ \frac{\gamma^{2/6 + 1/6}}{10} = \gamma^{3/2} $$ $$ \frac{\gamma^{3/6}}{10} = \gamma^{3/2} $$ $$ \frac{\gamma^{1/2}}{10} = \gamma^{3/2} $$ Since $\gamma$ is a positive real number, we can divide by $\gamma^{1/2}$: $$ \frac{1}{10} = \frac{\gamma^{3/2}}{\gamma^{1/2}} $$ $$ \frac{1}{10} = \gamma^{3/2 - 1/2} $$ $$ \frac{1}{10} = \gamma^1 $$ Thus, $\gamma = \frac{1}{10}$. Step 4: Calculate $\alpha^3$. Substitute $\gamma = \frac{1}{10}$ into the expression for $\alpha$: $$ \alpha = \frac{\gamma^{1/3}}{2} = \frac{(1/10)^{1/3}}{2} $$ Now, cube both sides: $$ \alpha^3 = \left(\frac{(1/10)^{1/3}}{2}\right)^3 = \frac{(1/10)^{(1/3) \cdot 3}}{2^3} = \frac{1/10}{8} = \frac{1}{80} $$ Step 5: Calculate $\beta^{12}$. Substitute $\gamma = \frac{1}{10}$ into the expression for $\beta$: $$ \beta = \frac{\gamma^{1/6}}{5} = \frac{(1/10)^{1/6}}{5} $$ Now, raise both sides to the power of 12: $$ \beta^{12} = \left(\frac{(1/10)^{1/6}}{5}\right)^{12} = \frac{(1/10)^{(1/6) \cdot 12}}{5^{12}} = \frac{(1/10)^2}{5^{12}} = \frac{1/100}{5^{12}} $$ $$ \beta^{12} = \frac{1}{100 \cdot 5^{12}} = \frac{1}{(10^2) \cdot 5^{12}} = \frac{1}{(2 \cdot 5)^2 \cdot 5^{12}} = \frac{1}{2^2 \cdot 5^2 \cdot 5^{12}} = \frac{1}{2^2 \cdot 5^{14}} = \frac{1}{4 \cdot 5^{14}} $$