What is $4
log _{\frac{1}{2}} w+\left(2
log _{\frac{1}{2}} u-3
log _{\frac{1}{2}} v\right)$ written as a single logarithm?
A. $
log _{\frac{1}{2}} w^4 u^2-v^3$
B. $
log _{\frac{1}{2}} w^4\left(\frac{u^2}{v^3}\right)$
C. $
log _{\frac{1}{2}}\left(\frac{w^4}{u^2 v^3}\right)$
D. $
log _{\frac{1}{2}}\left(w\left(\frac{u^2}{v^3}\right)\right)^4$
Answer (2)
The expression 4 lo g 2 1 w + ( 2 lo g 2 1 u − 3 lo g 2 1 v ) can be combined into a single logarithm as lo g 2 1 ( v 3 w 4 u 2 ) . Therefore, the correct answer is option C.
;
Apply the power rule of logarithms: a lo g b x = lo g b x a .
Apply the sum rule of logarithms: lo g b x + lo g b y = lo g b ( x y ) .
Apply the difference rule of logarithms: lo g b x − lo g b y = lo g b ( y x ) .
Combine the terms to get the final answer: lo g 2 1 v 3 w 4 u 2 .
Explanation
Understanding the Problem We are given the expression 4 lo g 2 1 w + ( 2 lo g 2 1 u − 3 lo g 2 1 v ) and we want to write it as a single logarithm. We will use the properties of logarithms to achieve this.
Applying the Power Rule First, we use the power rule of logarithms, which states that a lo g b x = lo g b x a . Applying this rule to each term, we get:
4 lo g 2 1 w = lo g 2 1 w 4 2 lo g 2 1 u = lo g 2 1 u 2 3 lo g 2 1 v = lo g 2 1 v 3
Substituting these back into the original expression, we have:
lo g 2 1 w 4 + ( lo g 2 1 u 2 − lo g 2 1 v 3 )
Applying the Sum Rule Next, we use the sum and difference rules of logarithms. The sum rule states that lo g b x + lo g b y = lo g b ( x y ) , and the difference rule states that lo g b x − lo g b y = lo g b ( y x ) .
Applying the sum rule to the first two terms, we get:
lo g 2 1 w 4 + lo g 2 1 u 2 = lo g 2 1 ( w 4 u 2 )
Now, we have:
lo g 2 1 ( w 4 u 2 ) − lo g 2 1 v 3
Applying the Difference Rule Finally, we apply the difference rule to combine the remaining terms:
lo g 2 1 ( w 4 u 2 ) − lo g 2 1 v 3 = lo g 2 1 v 3 w 4 u 2
Final Answer Therefore, the expression 4 lo g 2 1 w + ( 2 lo g 2 1 u − 3 lo g 2 1 v ) written as a single logarithm is lo g 2 1 v 3 w 4 u 2 .
Examples
Logarithms are incredibly useful in many real-world scenarios, especially when dealing with quantities that span a wide range of values. For instance, in chemistry, the pH scale uses logarithms to measure the acidity or alkalinity of a solution. Similarly, in seismology, the Richter scale uses logarithms to quantify the magnitude of earthquakes. In finance, logarithmic scales are often used to represent stock market indices or investment growth over time, making it easier to visualize and compare percentage changes. Understanding how to manipulate and combine logarithmic expressions, like we did in this problem, is essential for interpreting and working with these types of data effectively.
For example, if you are comparing the loudness of sounds, which are measured in decibels using a logarithmic scale, you might need to combine different sound sources to find the total loudness. If one sound has an intensity w , another has intensity u , and a third has intensity v , and you want to express the combined intensity as a single logarithmic value, you would use the properties we applied in this problem. The expression lo g 2 1 v 3 w 4 u 2 could represent a complex combination of these intensities, where w is amplified four times, u is amplified twice, and v is reduced three times relative to the base intensity.
