We are given the expression:

$\ln\left(x\sqrt{1+x^2}\right)$

Step-by-step solution:

1. Break the product inside the logarithm: Use the logarithmic property: $\ln(a \cdot b) = \ln(a) + \ln(b)$ Here, $a = x$ and $b = \sqrt{1+x^2}$, so: $\ln\left(x\sqrt{1+x^2}\right) = \ln(x) + \ln\left(\sqrt{1+x^2}\right)$

2. Simplify the square root term: Next, simplify the second logarithmic term. The square root can be written as an exponent: $\sqrt{1+x^2} = (1+x^2)^{1/2}$ Now, apply the logarithmic property: $\ln(a^n) = n\ln(a)$ So: $\ln\left(\sqrt{1+x^2}\right) = \frac{1}{2} \ln(1+x^2)$

3. Final expression: Now combine everything: $\ln\left(x\sqrt{1+x^2}\right) = \ln(x) + \frac{1}{2} \ln(1+x^2)$

Thus, the expression written as a sum of logarithms is:

$\ln(x) + \frac{1}{2} \ln(1+x^2)$

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Do you have any questions or would you like further details on any part?

Here are 5 related questions:

1. How would you simplify $\ln\left(\frac{a}{b}\right)$?
2. What is the logarithmic property for a power function like $\ln(a^n)$?
3. How can you expand $\ln(ab^2)$ using logarithmic properties?
4. What happens to the logarithmic properties if $x$ is negative?
5. How do we differentiate $\ln\left(x^2 + 1\right)$?

Tip: Always remember that logarithmic properties allow you to simplify products, quotients, and powers inside logarithms!