Question

Simplify1n e^7 - 3 1n e^1/2 + 1n 1 / 10 - 1/2 1n e ^2/3Round to two decimal places.

Answer 1

Given:

$\frac{\ln e^7-3\ln e^{\frac{1}{2}}+\ln 1}{10-\frac{1}{2}\ln e^{\frac{2}{3}}}$

Simplify the expression,

$\begin{gathered} \frac{\ln e^7-3\ln e^{\frac{1}{2}}+\ln1}{10-\frac{1}{2}\ln e^{\frac{2}{3}}} \\ \text{ Use the rule:ln}^{}a^b_{}=b\ln a \\ =\frac{7\ln e-3(\frac{1}{2})\ln e^{}+\ln 1}{10-\frac{1}{2}(\frac{2}{3})\ln e^{}} \\ =\frac{7-\frac{3}{2}+0}{10-\frac{1}{3}}\ldots\text{. Since }\ln \text{e=1,}\ln \text{1=0} \\ =\frac{\frac{11}{2}}{\frac{29}{3}} \\ =\frac{11}{2}\cdot\frac{3}{29} \\ =\frac{33}{58} \end{gathered}$

Answer:

$\frac{\ln e^7-3\ln e^{\frac{1}{2}}+\ln1}{10-\frac{1}{2}\ln e^{\frac{2}{3}}}=\frac{33}{58}=0.57$