Step 1: Convert all logarithms to a common base, which is base 2. We use the change of base formula: $\log_b a = \frac{\log_c a}{\log_c b}$. For $\log_4 x$: $$ \log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2} $$ For $\log_{16} x$: $$ \log_{16} x = \frac{\log_2 x}{\log_2 16} = \frac{\log_2 x}{4} $$ Step 2: Substitute these expressions back into the original equation. $$ \log_2 x + \frac{\log_2 x}{2} + \frac{\log_2 x}{4} = \frac{21}{4} $$ Step 3: Let $y = \log_2 x$ to simplify the equation. $$ y + \frac{y}{2} + \frac{y}{4} = \frac{21}{4} $$ Step 4: Combine the terms on the left side by finding a common denominator, which is 4. $$ \frac{4y}{4} + \frac{2y}{4} + \frac{y}{4} = \frac{21}{4} $$ $$ \frac{4y + 2y + y}{4} = \frac{21}{4} $$ $$ \frac{7y}{4} = \frac{21}{4} $$ Step 5: Solve for $y$. Multiply both sides by 4: $$ 7y = 21 $$ Divide by 7: $$ y = \frac{21}{7} $$ $$ y = 3 $$ Step 6: Substitute back $y = \log_2 x$ and solve for $x$. $$ \log_2 x = 3 $$ By the definition of a logarithm, if $\log_b a = c$, then $b^c = a$. $$ x = 2^3 $$ $$ x = 8 $$ The final answer is $\boxed{8}$. Send me the next one 📸