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log_(16)4=

$\log _{16} 4=$

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Relationship between squares and square roots

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Q.

\log _{16} 4=

Identify Base: In $\log_{16}4$ , $16$ is the base. $\newline$ Rewrite $4$ as a power of $16$ . $\newline$ Since $4$ is not a direct power of $16$ , we need to express both $4$ and $16$ in terms of a common base that is a factor of both numbers. The common base that can be used here is $2$ , because $4$ is $16$ $0$ and $16$ is $16$ $2$ . $\newline$ $16$ $3$ $\newline$ $16$ $4$
Rewrite Numbers as Powers: Now that we have both numbers as powers of $2$ , we can rewrite the logarithm in terms of base $2$ . $\newline$ $log_{(2^4)}2^2$ becomes $log_{(2^4)}2^2$ .
Convert to Base $2$ : Apply the logarithm power rule, which states that $\log_b(a^c) = c \cdot \log_b(a)$ , to simplify the expression. $\log_{2^4}2^2$ becomes $2 \cdot \log_{2^4}2$ .
Apply Logarithm Power Rule: Now, we can simplify the logarithm because the base of the logarithm $2^4$ and the number inside the logarithm $2$ have the same base. $\log_{2^4}2$ is asking ＂to what power do we raise $2^4$ to get $2$ ?＂ Since $2^4$ is $16$ and we want to get $2$ , we need to raise $16$ to the power of $1/4$ . $2$ $0$
Simplify Logarithm: Multiply the result from Step $4$ by the coefficient from Step $3$ . $\newline$ $2 \times \left(\frac{1}{4}\right)$ $\newline$ $2 \times \frac{1}{4} = \frac{1}{2}$

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