$10$ is a base number of an exponential term and its exponent is the sum of number $2$ and half of the common logarithm of $16$. The entire exponential term is placed under a square root. The value of this radical has to find in this problem.

$\sqrt{\large 10^{\displaystyle \normalsize 2+\frac{1}{2} \log_{10} 16}}$

The exponent contains a logarithm term and the logarithmic term is multiplied by rational number $\dfrac{1}{2}$. It can be cancelled if the number of the logarithmic term is expressed in exponential notation.

Therefore, write the number $16$ in exponential notation on the basis of $4$.

$= \,\,$ $\sqrt{\large 10^{\displaystyle \normalsize 2+\frac{1}{2} \log_{10} 4^2}}$

Use power rule of logarithms to shift the power of the exponential term of the logarithm.

$= \,\,$ $\sqrt{\large 10^{\displaystyle \normalsize 2+\frac{2}{2} \log_{10} 4}}$

$= \,\,$ $\sqrt{\large 10^{\displaystyle \normalsize 2+\require{cancel} \frac{\cancel{2}}{\cancel{2}} \log_{10} 4}}$

$= \,\,$ $\sqrt{\large 10^{\displaystyle \normalsize 2+ \log_{10} 4}}$

The radicand can be split as two multiplying factors in exponential notation by the product rule of exponents.

$= \,\,$ $\sqrt{\large 10^{\displaystyle \normalsize 2} \times \large 10^{\displaystyle \normalsize \log_{10} 4}}$

The value of second exponential term is $4$ as per the fundamental logarithm identity.

$= \,\,$ $\sqrt{\large 10^{\displaystyle \normalsize 2} \times 4}$

Now, simplify the radical to obtain the answer for this problem.

$= \,\,$ $\sqrt{\large 10^{\displaystyle \normalsize 2} \times 2^2}$

$= \,\,$ $\sqrt{\large 10^{\displaystyle \normalsize 2}} \times \sqrt{\large 2^2}$

$= \,\,$ $10 \times 2$

$= \,\,$ $20$

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