## College Algebra (6th Edition)

$\frac{1}{2}+\frac{1}{2}ln(x)$
We know that $ln(x)$ is a natural logarithm with an understood base of $e$. Therefore, $ln\sqrt (ex)=log_{e}\sqrt (ex)$. Based on the product rule of logarithms, we know that $log_{b}(MN)=log_{b}M+log_{b}N$ (for $M\gt0$ and $N\gt0$). Therefore, $log_{e}\sqrt (ex)=log_{e}\sqrt e+log_{e}\sqrt x=log_{e}e^{\frac{1}{2}}+log_{e}x^{\frac{1}{2}}$. According to the power rule of logarithms, we know that $log_{b}M^{p}=plog_{b}M$ (when $b$ and $M$ are positive real numbers, $b\ne1$, and $p$ is any real number). Therefore, $log_{e}e^{\frac{1}{2}}+log_{e}x^{\frac{1}{2}}=\frac{1}{2}log_{e}e+\frac{1}{2}log_{e}x$. Furthermore, based on the definition of the logarithmic function, we know that $y=log_{b}x$ is equivalent to $b^{y}=x$ (for $x\gt0$ and $b\gt0$, $b\ne1$). Therefore, $log_{e}e=1$, because $e^{1}=e$. So, $log_{e}e+\frac{1}{2}log_{e}x=\frac{1}{2}(1)+\frac{1}{2}log_{e}x=\frac{1}{2}+\frac{1}{2}log_{e}x=\frac{1}{2}+\frac{1}{2}ln(x)$.