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2log_(6)(33)~~

$2 \log _{6}(33) \approx$

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Evaluate logarithms

Full solution

Q.

2 \log _{6}(33) \approx

Identify base and number: Identify the base of the logarithm, which is $6$ in this case, and the number we are taking the log of, which is $33$ .
Use power rule: Use the power rule of logarithms which states that $a\log_b(c) = \log_b(c^a)$ . So, $2\log_{6}(33)$ becomes $\log_{6}(33^2)$ .
Calculate $33^2$ : Calculate $33^2$ to simplify the expression. $33^2 = 1089$ .
Express as power of $6$ : Now we have $\log_{6}(1089)$ . We need to express $1089$ as a power of $6$ to evaluate the logarithm.
Unable to simplify: Try to express $1089$ as a power of $6$ . However, $1089$ is not a power of $6$ , so we cannot simplify this logarithm to an integer.

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