Given that [tex]$x=\log 25$[/tex] and [tex]$y=\log x$[/tex], express [tex]$\log _2 12$[/tex] in terms of [tex]$x$[/tex] and [tex]$y$[/tex].
Answer (1)
Express lo g 2 12 using the change of base formula as l o g 2 l o g 12 .
Simplify lo g 12 to lo g 3 + 2 lo g 2 .
Express lo g 2 in terms of x as 1 − 2 x .
Express lo g 12 in terms of x and y as y + lo g 1.2 , leading to the final answer: 1 − 2 x y + lo g 1.2 .
Explanation
Understanding the Problem We are given that x = lo g 25 and y = lo g x . Our goal is to express lo g 2 12 in terms of x and y . We know that lo g refers to lo g 10 .
Change of Base First, let's express lo g 2 12 using the change of base formula: lo g 2 12 = lo g 2 lo g 12
Simplifying the Logarithm We can simplify lo g 12 as follows: lo g 12 = lo g ( 3 × 4 ) = lo g 3 + lo g 4 = lo g 3 + 2 lo g 2 So, lo g 2 12 = lo g 2 lo g 3 + 2 lo g 2 = lo g 2 lo g 3 + 2
Expressing log 2 in terms of x Now, let's express lo g 2 in terms of x . We know that x = lo g 25 = lo g ( 5 2 ) = 2 lo g 5 . Therefore, lo g 5 = 2 x . Also, we know that lo g 2 = lo g 5 10 = lo g 10 − lo g 5 = 1 − lo g 5 = 1 − 2 x .
Rewriting log_2 12 We are given y = lo g x . So, 1 0 y = x . We need to find lo g 3 in terms of x and y . We have lo g 2 = 1 − 2 x . We can rewrite lo g 2 12 as: lo g 2 12 = lo g 2 lo g 12 = 1 − 2 x lo g 12
Expressing log 12 in terms of x and y Now, let's express lo g 12 in terms of x and y . We have lo g 12 = lo g ( 3 × 4 ) = lo g 3 + 2 lo g 2 . We need to find lo g 3 . We know that x = 1 0 y , so lo g x = y . We also know that x = lo g 25 , so 25 = 1 0 x . Therefore, lo g 5 = 2 x and lo g 2 = 1 − 2 x .
We can write lo g 12 = lo g ( l o g x x × 1.2 ) = lo g x + lo g 1.2 = y + lo g 1.2 .
Final Expression Substituting these expressions into the equation for lo g 2 12 , we get: lo g 2 12 = 1 − 2 x y + lo g 1.2
Final Answer Therefore, lo g 2 12 in terms of x and y is: 1 − 2 x y + lo g 1.2
Examples
Imagine you're a sound engineer adjusting audio frequencies. The logarithm base 2, or lo g 2 , is often used to describe octaves in audio. If you need to shift a sound's frequency and you know the original frequency's logarithm (base 10) and some related logarithmic values, you can use the formula we derived to find the new frequency's octave relative to a reference. This kind of calculation helps in audio mixing and mastering to fine-tune sound characteristics. Understanding logarithmic relationships allows precise control over audio properties, ensuring optimal sound quality and balance.
