2.303
1) For the following expression:
[tex]\frac{\ln30+\ln15}{\log_{10}30+\log_{10}15}[/tex]
We can simplify that and then round it off to the nearest thousandth:
2) Let's rewrite them simplifying using the logarithm property of multiplication:
[tex]\begin{gathered} \frac{\ln30+\ln15}{\log_{10}30+\log_{10}15}= \\ \frac{\ln(30\cdot15)}{\log_{10}30+\log_{10}15}= \\ \frac{\ln(30\cdot15)}{\log_{10}(30\cdot15)}= \\ \frac{\ln(450)}{\log_{10}(450)}= \end{gathered}[/tex]
Note that the base of the Natural Log is the Euler's number "e" so let's move on now using the calculator, finally:
[tex]\frac{\ln(450)}{\log_{10}(450)}=\frac{6.10924}{2.65321}=2.30258\ldots\approx2.303[/tex]
Note that only at the last step we have rounded it off. And that's the
answer
