Chapter 14: Problem 1
Evaluate (a) \(\log _{3} 9\) (b) \(\log _{10} 10\) (c) \(\log _{16} 8\)
Short Answer
Expert verified
(a) 2, (b) 1, (c) \( \frac{3}{4} \).
Step by step solution
01
Understanding Logarithms
The logarithm \( \log_b a \) is the exponent to which the base \( b \) must be raised to yield the number \( a \). For example, if \( b^n = a \), then \( \log_b a = n \).
02
Evaluating \( \log_3 9 \)
We need to find \( n \) such that \( 3^n = 9 \). Since \( 3^2 = 9 \), it follows that \( \log_3 9 = 2 \).
03
Evaluating \( \log_{10} 10 \)
The base \( 10 \) must be raised to what power to get \( 10 \)? Since \( 10^1 = 10 \), it follows that \( \log_{10} 10 = 1 \).
04
Evaluating \( \log_{16} 8 \) using Change of Base Formula
We don't have an exact power of \( 16 \) that gives \( 8 \), so we use the change of base formula: \( \log_{16} 8 = \frac{\log_{10} 8}{\log_{10} 16} \). We know that \( 8 = 2^3 \) and \( 16 = 2^4 \), so, \( \log_{16} 8 = \frac{3 \log_{10} 2}{4 \log_{10} 2} = \frac{3}{4} \).
05
State the Result
(a) \( \log_3 9 = 2 \), (b) \( \log_{10} 10 = 1 \), and (c) \( \log_{16} 8 = \frac{3}{4} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Exponents
Exponents are a way of expressing repeated multiplication of the same number. For example, the expression \( 3^2 \) means \( 3 \) multiplied by itself, which equals 9. In general, when you see a number like \( a^b \), the \( b \) is the exponent and it tells us how many times to multiply \( a \) by itself.
Exponents have several key properties that are good to understand:
Exponents have several key properties that are good to understand:
- Multiplying Powers: \( a^m \times a^n = a^{m+n} \)
- Dividing Powers: \( \frac{a^m}{a^n} = a^{m-n} \)
- Power of a Power: \( (a^m)^n = a^{m \times n} \)
Change of Base Formula
The change of base formula is a useful technique in logarithms that allows us to calculate a logarithm with any base by converting it to a fraction of two logarithms with a new, common base. If you're dealing with logarithms that don't result in nice whole numbers or easily recognizable values, the change of base formula becomes very handy.
The formula is:
For example, we used it to solve \( \log_{16} 8 \) in the exercise. By expressing both 8 and 16 as powers of 2 (\( 8 = 2^3 \) and \( 16 = 2^4 \)), we were able to simplify the problem using the change of base formula. This process showed that the answer is \( \frac{3}{4} \).
The formula is:
- \( \log_b a = \frac{\log_c a}{\log_c b} \)
For example, we used it to solve \( \log_{16} 8 \) in the exercise. By expressing both 8 and 16 as powers of 2 (\( 8 = 2^3 \) and \( 16 = 2^4 \)), we were able to simplify the problem using the change of base formula. This process showed that the answer is \( \frac{3}{4} \).
Logarithmic Functions
Logarithmic functions are the inverse of exponential functions. This means that while exponential functions take a base number and an exponent to produce a result, logarithmic functions take a result and a base to find the exponent. If you know that \( b^n = a \), then \( \log_b a = n \).
Logarithmic functions have several important properties:
Logarithmic functions have several important properties:
- Product Rule: \( \log_b(xy) = \log_b x + \log_b y \)
- Quotient Rule: \( \log_b\left(\frac{x}{y}\right) = \log_b x - \log_b y \)
- Power Rule: \( \log_b(x^a) = a \times \log_b x \)