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^{(1)/(2)}+\ln 1}{10-(1)/(2)\ln e^{(2)/(3)}}`

Simplify the expression,

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

Answer:

`\frac{\ln e^7-3\ln e^{(1)/(2)}+\ln1}{10-(1)/(2)\ln e^{(2)/(3)}}=(33)/(58)=0.57`