Question

$$0^{-12}w/m^{2}; \log_{5}125+\log_{e}\sqrt{e}=?$$

Answer

$$INF*w/m^2;log(_5,125)+1/2$$

Solution


Use Power Rule: \(\ln{({x}^{y})}=y\ln{x}\).
\[\begin{aligned}&{0}^{-12}\times \frac{w}{{m}^{2}}\\&\log_{_5}{125}+\frac{1}{2}\ln{e}\end{aligned}\]
Use Rule of e: \(\ln{e}=1\).
\[\begin{aligned}&{0}^{-12}\times \frac{w}{{m}^{2}}\\&\log_{_5}{125}+\frac{1}{2}\times 1\end{aligned}\]
Use Negative Power Rule: \({x}^{-a}=\frac{1}{{x}^{a}}\).
\[\begin{aligned}&\frac{1}{{0}^{12}}\times \frac{w}{{m}^{2}}\\&\log_{_5}{125}+\frac{1}{2}\times 1\end{aligned}\]
Simplify  \({0}^{12}\)  to  \(0\).
\[\begin{aligned}&\frac{1}{0}\times \frac{w}{{m}^{2}}\\&\log_{_5}{125}+\frac{1}{2}\times 1\end{aligned}\]
Simplify  \(\frac{1}{0}\)  to  \(\infty \).
\[\begin{aligned}&\infty \times \frac{w}{{m}^{2}}\\&\log_{_5}{125}+\frac{1}{2}\times 1\end{aligned}\]
Simplify  \(\frac{1}{2}\times 1\)  to  \(\frac{1}{2}\).
\[\begin{aligned}&\infty \times \frac{w}{{m}^{2}}\\&\log_{_5}{125}+\frac{1}{2}\end{aligned}\]