Given that [tex]$\log _a(2) \approx 0.5$[/tex] and [tex]$\log _a(5) \approx 1.16$[/tex], evaluate each of the following. Hint: use the properties of logarithms to rewrite the given logarithm in terms of the logarithms of 2 and 5.
a) [tex]$\log _a(0.4) \approx$[/tex] $\square$
b) [tex]$\log _a(\sqrt{5}) \approx$[/tex] $\square$
c) [tex]$\log _a(16) \approx$[/tex] $\square$
d) [tex]$\log _a(50) \approx$[/tex] $\square$
e) [tex]$\log _a(10) \approx$[/tex] $\square$
f) [tex]$\log _a(0.8) \approx$[/tex] $\square$
Answer (2)
Rewrite each logarithm in terms of lo g a ( 2 ) and lo g a ( 5 ) using logarithm properties.
Substitute the given approximations lo g a ( 2 ) ≈ 0.5 and lo g a ( 5 ) ≈ 1.16 .
Calculate the approximate values:
lo g a ( 0.4 ) = lo g a ( 2/5 ) = lo g a ( 2 ) − lo g a ( 5 ) ≈ − 0.66
lo g a ( 5 ) = ( 1/2 ) lo g a ( 5 ) ≈ 0.58
lo g a ( 16 ) = 4 lo g a ( 2 ) ≈ 2
lo g a ( 50 ) = lo g a ( 2 ⋅ 5 2 ) = lo g a ( 2 ) + 2 lo g a ( 5 ) ≈ 2.82
lo g a ( 10 ) = lo g a ( 2 ⋅ 5 ) = lo g a ( 2 ) + lo g a ( 5 ) ≈ 1.66
lo g a ( 0.8 ) = lo g a ( 8/10 ) = lo g a ( 2 2 /5 ) = 2 lo g a ( 2 ) − lo g a ( 5 ) ≈ − 0.16
State the final answers: lo g a ( 0.4 ) ≈ − 0.66 , lo g a ( 5 ) ≈ 0.58 , lo g a ( 16 ) ≈ 2 , lo g a ( 50 ) ≈ 2.82 , lo g a ( 10 ) ≈ 1.66 , lo g a ( 0.8 ) ≈ − 0.16
Explanation
Understanding the Problem and Key Properties We are given that lo g a ( 2 ) ≈ 0.5 and lo g a ( 5 ) ≈ 1.16 . We will use the properties of logarithms to rewrite the given expressions in terms of lo g a ( 2 ) and lo g a ( 5 ) . The key properties are:
lo g a ( x y ) = lo g a ( x ) + lo g a ( y )
lo g a ( x / y ) = lo g a ( x ) − lo g a ( y )
lo g a ( x n ) = n lo g a ( x )
Evaluating log_a(0.4) a) We have lo g a ( 0.4 ) . Since 0.4 = 10 4 = 5 2 , we can rewrite the expression as: lo g a ( 0.4 ) = lo g a ( 5 2 ) = lo g a ( 2 ) − lo g a ( 5 ) ≈ 0.5 − 1.16 = − 0.66 So, lo g a ( 0.4 ) ≈ − 0.66 .
Evaluating log_a(sqrt(5)) b) We have lo g a ( 5 ) . Since 5 = 5 2 1 , we can rewrite the expression as: lo g a ( 5 ) = lo g a ( 5 2 1 ) = 2 1 lo g a ( 5 ) ≈ 2 1 ( 1.16 ) = 0.58 So, lo g a ( 5 ) ≈ 0.58 .
Evaluating log_a(16) c) We have lo g a ( 16 ) . Since 16 = 2 4 , we can rewrite the expression as: lo g a ( 16 ) = lo g a ( 2 4 ) = 4 lo g a ( 2 ) ≈ 4 ( 0.5 ) = 2 So, lo g a ( 16 ) ≈ 2 .
Evaluating log_a(50) d) We have lo g a ( 50 ) . Since 50 = 2 ⋅ 5 2 , we can rewrite the expression as: lo g a ( 50 ) = lo g a ( 2 ⋅ 5 2 ) = lo g a ( 2 ) + 2 lo g a ( 5 ) ≈ 0.5 + 2 ( 1.16 ) = 0.5 + 2.32 = 2.82 So, lo g a ( 50 ) ≈ 2.82 .
Evaluating log_a(10) e) We have lo g a ( 10 ) . Since 10 = 2 ⋅ 5 , we can rewrite the expression as: lo g a ( 10 ) = lo g a ( 2 ⋅ 5 ) = lo g a ( 2 ) + lo g a ( 5 ) ≈ 0.5 + 1.16 = 1.66 So, lo g a ( 10 ) ≈ 1.66 .
Evaluating log_a(0.8) f) We have lo g a ( 0.8 ) . Since 0.8 = 10 8 = 5 4 = 5 2 2 , we can rewrite the expression as: lo g a ( 0.8 ) = lo g a ( 5 2 2 ) = 2 lo g a ( 2 ) − lo g a ( 5 ) ≈ 2 ( 0.5 ) − 1.16 = 1 − 1.16 = − 0.16 So, lo g a ( 0.8 ) ≈ − 0.16 .
Final Answer In summary: a) lo g a ( 0.4 ) ≈ − 0.66 b) lo g a ( 5 ) ≈ 0.58 c) lo g a ( 16 ) ≈ 2 d) lo g a ( 50 ) ≈ 2.82 e) lo g a ( 10 ) ≈ 1.66 f) lo g a ( 0.8 ) ≈ − 0.16
Examples
Logarithms are incredibly useful in many real-world applications. For instance, they are used in calculating the magnitude of earthquakes on the Richter scale, which is a base-10 logarithmic scale. Similarly, in chemistry, pH values, which measure the acidity or alkalinity of a solution, are also based on a logarithmic scale. Understanding how to manipulate and evaluate logarithms allows us to work with and interpret these scales effectively, providing insights into phenomena ranging from seismic activity to chemical reactions. Logarithmic scales help compress a wide range of values into a more manageable and interpretable format.
Using the given approximations for lo g a ( 2 ) and lo g a ( 5 ) , we calculated the values of various logarithms by applying logarithmic properties. The approximations are as follows: lo g a ( 0.4 ) ≈ − 0.66 , lo g a ( 5 ) ≈ 0.58 , lo g a ( 16 ) ≈ 2 , lo g a ( 50 ) ≈ 2.82 , lo g a ( 10 ) ≈ 1.66 , and lo g a ( 0.8 ) ≈ − 0.16 .
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