If ${\log }_2(x)+{\log }_4(x)+{\log }_{16}(x)=7$ for a positive real number $x$, find the value of $x$.

Step 1: Express all logarithms in base $2$ using the base change formula ${\log }_{a^k}(x)=\frac{1}{k}{\log }_a(x)$:

• ${\log }_4(x)={\log }_{2^2}(x)=\frac{1}{2}{\log }_2(x)$

• ${\log }_{16}(x)={\log }_{2^4}(x)=\frac{1}{4}{\log }_2(x)$

Step 2: Substitute these into the given equation:

Step 3: Group the coefficients:

Step 4: Solve for ${\log }_2(x)$:

Step 5: Solve for $x$:

So the value of $x$ is $16$.