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Which of the following is equivalent to 
(log_(9)(m))/(log(m)) ?
Choose 1 answer:
(A) 
log(9)
(B) 
log_(9)(1)
(c) 
(1)/(log(m))
(D) 
(1)/(log(9))

Which of the following is equivalent to log9(m)log(m) \frac{\log _{9}(m)}{\log (m)} ?\newlineChoose 11 answer:\newline(A) log(9) \log (9) \newline(B) log9(1) \log _{9}(1) \newline(C) 1log(m) \frac{1}{\log (m)} \newline(D) 1log(9) \frac{1}{\log (9)}

Full solution

Q. Which of the following is equivalent to log9(m)log(m) \frac{\log _{9}(m)}{\log (m)} ?\newlineChoose 11 answer:\newline(A) log(9) \log (9) \newline(B) log9(1) \log _{9}(1) \newline(C) 1log(m) \frac{1}{\log (m)} \newline(D) 1log(9) \frac{1}{\log (9)}
  1. Recognize logarithm property: Recognize the property of logarithms that can be used to simplify the expression.\newlineThe expression (log9(m))/(log(m))(\log_{9}(m))/(\log(m)) can be simplified using the change of base formula for logarithms.\newlineThe change of base formula states that logb(a)=logc(a)/logc(b)\log_b(a) = \log_c(a) / \log_c(b), where bb and cc are bases and aa is the argument of the logarithm.
  2. Apply change of base formula: Apply the change of base formula to the given expression.\newlineUsing the change of base formula, we can rewrite (log9(m))/(log(m))(\log_{9}(m))/(\log(m)) as:\newline(log9(m))/(log(m))=1/(log(m)/log(9))(\log_{9}(m))/(\log(m)) = 1 / (\log(m) / \log(9))\newlineThis is because log9(m)\log_{9}(m) is equivalent to log(m)/log(9)\log(m) / \log(9) by the change of base formula.
  3. Simplify the expression: Simplify the expression.\newlineSimplifying the expression, we get:\newline1(log(m)log(9))=log(9)log(m)\frac{1}{\left(\frac{\log(m)}{\log(9)}\right)} = \frac{\log(9)}{\log(m)}\newlineThis simplifies to:\newline(log(9)log(m))=1(log(m)log(9))\left(\frac{\log(9)}{\log(m)}\right) = \frac{1}{\left(\frac{\log(m)}{\log(9)}\right)}
  4. Identify correct answer: Identify the correct answer from the given options.\newlineThe simplified expression 1(log(m)log(9)) \frac{1}{\left(\frac{\log(m)}{\log(9)}\right)} matches option (D) 1log(9) \frac{1}{\log(9)} .

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