Condense the logarithm k log a+g log d Answer: log(◻)

Rewrite terms as logarithms: We are given the expression k log a + g log d and we need to condense it into a single logarithm. According to the properties of logarithms, specifically the power rule, which states that m log_b(n) = log_b(n^m), we can rewrite each term as a logarithm of a power. For the first term, k log a becomes log(a^k). For the second term, g log d becomes log(d^g).Combine logarithms using product rule: Now that we have rewritten the terms, we can use another property of logarithms, the product rule, which states that log_b(m) + log_b(n) = log_b(m * n), to combine the two logarithms into one. So, log(a^k) + log(d^g) becomes log(a^k * d^g).Final condensed expression: We have now condensed the original expression into a single logarithm. The final answer is log(a^k * d^g).

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Condense the logarithm

k log a+g log d
Answer:
log(◻)

Condense the logarithm $\newline$ $k \log a+g \log d$ $\newline$ Answer: $\log (\square)$

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Precalculus

Convert between exponential and logarithmic form

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Q. Condense the logarithm

\newline

k \log a+g \log d

\newline

Answer:

\log (\square)

Rewrite terms as logarithms: We are given the expression $k \log a + g \log d$ and we need to condense it into a single logarithm. $\newline$ According to the properties of logarithms, specifically the power rule, which states that $m \log_b(n) = \log_b(n^m)$ , we can rewrite each term as a logarithm of a power. $\newline$ For the first term, $k \log a$ becomes $\log(a^k)$ . $\newline$ For the second term, $g \log d$ becomes $\log(d^g)$ .
Combine logarithms using product rule: Now that we have rewritten the terms, we can use another property of logarithms, the product rule, which states that $\log_b(m) + \log_b(n) = \log_b(m \cdot n)$ , to combine the two logarithms into one. $\newline$ So, $\log(a^k) + \log(d^g)$ becomes $\log(a^k \cdot d^g)$ .
Final condensed expression: We have now condensed the original expression into a single logarithm. The final answer is $\log(a^k \cdot d^g)$ .