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${:[log_(3)8~~1.9],[log_(3)10~~2.1],[log_(3)11~~2.2],[" Find "log_(3)100]:}$

$22$ ) $\begin{array}{l}\log _{3} 8 \approx 1.9 \\ \log _{3} 10 \approx 2.1 \\ \log _{3} 11 \approx 2.2 \\ \text { Find } \log _{3} 100\end{array}$

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Find derivatives of logarithmic functions

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Q.

22

)

\begin{array}{l}\log _{3} 8 \approx 1.9 \\ \log _{3} 10 \approx 2.1 \\ \log _{3} 11 \approx 2.2 \\ \text { Find } \log _{3} 100\end{array}

Express $100$ as $10\times10$ : We are given approximate values for $\log_{3}(8)$ , $\log_{3}(10)$ , and $\log_{3}(11)$ . We need to find the value of $\log_{3}(100)$ . We can use the property of logarithms that states $\log_{b}(m\times n) = \log_{b}(m) + \log_{b}(n)$ , where $b$ is the base, $m$ , and $n$ are the numbers. We can express $100$ as $10\times10$ $1$ .
Apply logarithm property: Using the given approximations, we know that $\log_{3} 10$ is approximately $2.1$ . Since $100$ is $10 \times 10$ , we can write $\log_{3} 100$ as $\log_{3} (10 \times 10)$ .
Substitute approximate values: Applying the logarithm property mentioned earlier, we get $\log_{3} 100 = \log_{3} 10 + \log_{3} 10$ .
Calculate final value: Substituting the approximate value for $\log_{3} 10$ , we get $\log_{3} 100 = 2.1 + 2.1$ .
Calculate final value: Substituting the approximate value for $\log_{3} 10$ , we get $\log_{3} 100 = 2.1 + 2.1$ . Adding the two values together, we find that $\log_{3} 100$ is approximately $2.1 + 2.1 = 4.2$ .

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