Which expression is equivalent to $\log _3 \frac{c}{9}$ ?
$\log _3 c+\log _3(9)$
$\log _3(9)+\log _3(c)$
$\log _3 c-\log _3(9)$
$\log _3(9)-\log _3(c)$
Answer (1)
Apply the quotient rule of logarithms: lo g 3 9 c = lo g 3 c − lo g 3 9 .
Simplify lo g 3 9 to 2 , but keep the expression in terms of logarithms for comparison with the options.
The equivalent expression is lo g 3 c − lo g 3 ( 9 ) .
Therefore, the final answer is lo g 3 c − lo g 3 ( 9 ) .
Explanation
Understanding the problem We are given the expression lo g 3 9 c and asked to find an equivalent expression from the given options. We will use the properties of logarithms to rewrite the given expression.
Applying the Quotient Rule The quotient rule of logarithms states that lo g b ( y x ) = lo g b ( x ) − lo g b ( y ) . Applying this rule to our expression, we get: lo g 3 9 c = lo g 3 c − lo g 3 9
Simplifying the expression We can simplify lo g 3 9 further. Since 9 = 3 2 , we have lo g 3 9 = lo g 3 ( 3 2 ) = 2 . Therefore, the expression becomes: lo g 3 c − 2 However, we are looking for an expression in terms of lo g 3 c and lo g 3 9 . So, we leave it as: lo g 3 c − lo g 3 9
Finding the equivalent expression Comparing this result with the given options, we see that the equivalent expression is lo g 3 c − lo g 3 ( 9 ) .
Examples
Logarithms are used in many real-world applications, such as measuring the intensity of earthquakes (the Richter scale), the loudness of sounds (decibels), and the acidity of a solution (pH). The properties of logarithms, like the quotient rule used in this problem, help simplify complex calculations in these fields. For example, if you want to compare the intensities of two earthquakes, you can use the quotient rule to find the difference in their magnitudes on the Richter scale. If one earthquake is 100 times more intense than another, the difference in their Richter scale readings would be lo g 10 ( 100 ) = 2 .
