To solve the equation, express all logarithms with the common base $2$ using the property $\log_{b^k} a = \frac{1}{k} \log_b a$:

• $\log_4 x = \log_{2^2} x = \frac{1}{2} \log_2 x$
• $\log_{16} x = \log_{2^4} x = \frac{1}{4} \log_2 x$

Substituting these back into the equation: $\log_2 x + \frac{1}{2} \log_2 x + \frac{1}{4} \log_2 x = 7$ $\log_2 x \left(1 + \frac{1}{2} \frac{1}{4}\right) = 7$ $\log_2 x \left(\frac{7}{4}\right) = 7$ $\log_2 x = 4 \implies x = 2^4 = 16$.

Since $16 > 0$, it is a valid solution.