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Chapter 5: Problem 55

Given that \(\log _{a} 2=0.301, \log _{a} 7=0.845,\) and \(\log _{a} 11=1.041,\) find each of the following, if possible. Round the answer to the nearest thousandth. $$\log _{a} 98$$

Short Answer

Expert verified
1.991

Step by step solution

01

Factorize the given number

Express 98 as a product of its prime factors. Notice that 98 = 2 脳 7 脳 7 (or 2 脳 7^2).
02

Apply logarithm properties

Using the logarithm property \(\text{log}_a (x \times y) = \text{log}_a x + \text{log}_a y\), we can write \(\text{log}_a 98\) as \(\text{log}_a (2 \times 7^2)\).
03

Split the logarithms

Using the properties of logarithms, split \(\text{log}_a (2 \times 7^2)\) into \(\text{log}_a 2 + \text{log}_a 7^2\).
04

Apply the power rule

Using the power rule of logarithms \(\text{log}_a (y^k) = k \times \text{log}_a y\), rewrite \(\text{log}_a 7^2\) as \((2 \times \text{log}_a 7)\).
05

Substitute known values

Substitute the known values of \(\text{log}_a 2 = 0.301\) and \(\text{log}_a 7 = 0.845\) into the equation. You get \(\text{log}_a 2 + 2 \times \text{log}_a 7 = 0.301 + 2 \times 0.845\).
06

Perform the calculations

Complete the calculation \(\text{log}_a 98 = 0.301 + 2 \times 0.845 = 0.301 + 1.69 = 1.991\).
07

Round to the nearest thousandth

Round the result to the nearest thousandth. \(\text{log}_a 98 = 1.991\) is already rounded to the nearest thousandth.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

logarithm properties
Logarithms are a way to express exponentiation. They help in transforming multiplicative relationships into additive ones. The basic properties of logarithms simplify complex calculations. Here are some key properties:
  • \text{Product Property}: \(\text{log}_a (xy) = \text{log}_a x + \text{log}_a y\) This property allows you to split the logarithm of a product into the sum of the logarithms of individual factors.
  • \text{Quotient Property}: \(\text{log}_a (x/y) = \text{log}_a x - \text{log}_a y\) This property helps you break down the logarithm of a fraction.
  • \text{Power Rule}: \(\text{log}_a (x^k) = k \times \text{log}_a x\) This property is useful for handling exponents inside a logarithm.
These properties are pivotal in solving logarithmic problems. Always remember to apply them carefully in each specific scenario.
prime factorization
Prime factorization is the process of breaking down a composite number into its prime factors. Prime numbers are numbers greater than 1 that have no other divisors other than 1 and themselves.
For example, the prime factorization of 98 goes like this:
  • First, we divide 98 by the smallest prime number which is 2:$$ 98 \div 2 = 49$$Therefore, one factor is 2.
  • Next, we recognize that 49 is not prime, so we continue dividing by the next smallest prime number, which is 7: $$ 49 = 7 \times 7$$
  • So, we can say:$$ 98 = 2 \times 7^2$$
Prime factorization is crucial in simplifying logarithmic expressions, allowing us to use the properties of logarithms for calculation.
power rule of logarithms
The power rule of logarithms states that the logarithm of a number raised to an exponent can be expressed by multiplying the exponent by the logarithm of the base number. Mathematically, it is written as:
\text{log}_a (y^k) = k \times \text{log}_a y
This rule simplifies expressions where the argument of the logarithm is an exponential term. Here鈥檚 a step-by-step example:
  • Suppose you have \text{log}_a 7^2. Using the power rule:$$ \text{log}_a 7^2 = 2 \times \text{log}_a 7$$
  • If \text{log}_a 7 is already known, say 0.845, then we substitute the value in:$$ 2 \times 0.845 = 1.69$$
This rule is powerful in calculations involving logarithms and helps to break down and solve complex problems efficiently.

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Most popular questions from this chapter

Advertising. A company begins an Internet advertising campaign to market a new telephone. The percentage of the target market that buys a product is generally a function of the length of the advertising campaign. The estimated percentage is given by $$ f(t)=100\left(1-e^{-0.04 t}\right) $$ where \(t\) is the number of days of the campaign. a) Graph the function. b) Find \(f(25),\) the percentage of the target market that has bought the phone after a 25 -day advertising campaign. c) After how long will \(90 \%\) of the target market have bought the phone?

Determine whether each of the following is true or false. Assume that \(a, x, M,\) and \(N\) are positive. $$\frac{\log _{a} M}{\log _{a} N}=\log _{a} M-\log _{a} N$$

Solve using any method. $$\left|2^{x^{2}}-8\right|=3$$

Compound Interest. Suppose that \(\$ 82.000\) is invested at \(4 \frac{1}{2} \%\) interest, compounded quarterly. a) Find the function for the amount to which the investment grows after \(t\) years. b) Graph the function. c) Find the amount of money in the account at \(t=0,2\) \(5,\) and 10 years. d) When will the amount of money in the account reach \(\$ 100,000 ?\)

Express as a single logarithm and, if possible, simplify. $$\log _{a}\left(a^{10}-b^{10}\right)-\log _{a}(a+b)$$
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