The expression shown in the image is:

$5 \log 16 \cdot x^{2 \log \sqrt{3}}, 9 \log 5$

Simplifying the components:

1. First part: $5 \log 16 \cdot x^{2 \log \sqrt{3}}$

   • $5 \log 16$ is a constant multiplication term. We know that $\log 16 = \log (2^4) = 4 \log 2$. Thus: $5 \log 16 = 5 \cdot 4 \log 2 = 20 \log 2$
   • Now, simplify $x^{2 \log \sqrt{3}}$. Since $\sqrt{3} = 3^{1/2}$, we can express $\log \sqrt{3} = \frac{1}{2} \log 3$. Therefore: $x^{2 \log \sqrt{3}} = x^{2 \cdot \frac{1}{2} \log 3} = x^{\log 3}$ Using the logarithmic property $x^{\log a} = a^{\log x}$, we can simplify: $x^{\log 3} = 3^{\log x}$

   Hence, the entire first part simplifies to: $20 \log 2 \cdot 3^{\log x}$

2. Second part: $9 \log 5$

   This is simply: $9 \log 5$

Thus, the simplified form of the expression is:

$20 \log 2 \cdot 3^{\log x}, 9 \log 5$

Would you like more details on logarithmic properties or have any questions?

Related Questions:

1. How do you apply the power rule of logarithms in other scenarios?
2. Can you solve equations involving logarithmic exponents?
3. How would you solve for $x$ if given $3^{\log x} = a$?
4. What is the change of base formula for logarithms?
5. How do you evaluate logarithmic expressions involving different bases?

Tip:

Remember, logarithmic properties such as $\log (a^b) = b \log a$ are incredibly useful for simplifying complex expressions.