Write the logarithm as a sum and/or difference of logarithmic expressions. Eliminate exponents and radicals, and evaluate logarithms wherever possible. Assume that \( x, y>0 \). \( \ln \left(\frac{\sqrt{x^{8}+y}}{e^{3}}\right)= \)

To tackle the expression \( \ln \left(\frac{\sqrt{x^{8}+y}}{e^{3}}\right) \), we first apply the properties of logarithms. The logarithm of a quotient is the difference of the logarithms, so we can write: \[ \ln \left(\frac{\sqrt{x^{8}+y}}{e^{3}}\right) = \ln(\sqrt{x^{8}+y}) - \ln(e^{3}). \] Next, we know that \( \ln(e^{3}) = 3 \) since the natural logarithm of \( e \) raised to a power is just that power. Thus, it simplifies to: \[ \ln(\sqrt{x^{8}+y}) - 3. \] Now, let’s rewrite \( \ln(\sqrt{x^{8}+y}) \). We can express the square root as an exponent: \[ \sqrt{x^{8}+y} = (x^{8}+y)^{1/2}. \] Using the property of logarithms that states \( \ln(a^b) = b \ln(a) \), we get: \[ \ln(\sqrt{x^{8}+y}) = \frac{1}{2} \ln(x^{8}+y). \] Putting it all together, we have: \[ \ln \left(\frac{\sqrt{x^{8}+y}}{e^{3}}\right) = \frac{1}{2} \ln(x^{8}+y) - 3. \] This is the final expression, showing the logarithm as a sum and/or difference without exponents and radicals!