Questions: Taking logarithms and antilogarithms is necessary to solve many chemistry problems. For practice, complete the following table, where N is a number. N ln N log N --------- 6.18 □ □ 3.405 -0.380

Taking logarithms and antilogarithms is necessary to solve many chemistry problems. For practice, complete the following table, where N is a number.

N ln N log N
---------
6.18 □
□ 3.405
-0.380

Transcript text: Taking logarithms and antilogarithms is necessary to solve many chemistry problems. For practice, complete the following table, where $N$ is a number. \begin{tabular}{|l|l|l|} \hline \multicolumn{1}{|c|}{$N$} & $\ln N$ & $\log N$ \\ \hline 6.18 & & $\square$ \\ \hline$\square$ & 3.405 & \\ \hline & & -0.380 \\ \hline \end{tabular}

Solution

Solution Steps

Step 1: Calculating $\ln N$ and $\log N$ given $N$

Given $N = 6.18$, we calculate $\ln N$ using $\ln N = \log_e N$. Thus, $\ln N = 1.82$. Next, we calculate $\log N$ using $\log N = \log_{10} N$. Thus, $\log N = 0.79$.

Final Answer:

The completed table entries are: $N = 6.18$, $\ln N = 1.82$, and $\log N = 0.79$.

Step 1: Calculating $N$ and $\log N$ given $\ln N$

Given $\ln N = 3.405$, we calculate $N$ using $N = e^{\ln N}$. Thus, $N = 30.11$. Next, we calculate $\log N$ from $\ln N$ using $\log N = \ln N / \ln 10$. Thus, $\log N = 1.48$.

Final Answer:

The completed table entries are: $N = 30.11$, $\ln N = 3.4$, and $\log N = 1.48$.

Step 1: Calculating $N$ and $\ln N$ given $\log N$

Given $\log N = -0.38$, we calculate $N$ using $N = 10^{\log N}$. Thus, $N = 0.42$. Next, we calculate $\ln N$ from $\log N$ using $\ln N = \log N \cdot \ln 10$. Thus, $\ln N = -0.87$.

Final Answer:

The completed table entries are: $N = 0.42$, $\ln N = -0.87$, and $\log N = -0.38$.

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