/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Write the expression as a single... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Write the expression as a single logarithm with a coefficient of \(1 .\) $$\log _{10} 30+\log _{10} 2$$

Short Answer

Expert verified
\(\log_{10} 60\)

Step by step solution

01

Identify the Property of Logarithms

Recognize that when you have a sum of logarithms with the same base, such as \( \log_b M + \log_b N \), you can combine them using the Product Rule. This rule states that \( \log_b M + \log_b N = \log_b (M \cdot N) \).
02

Apply the Product Rule

Apply the Product Rule to the expression \( \log_{10} 30 + \log_{10} 2 \). Thus, this becomes \( \log_{10} (30 \times 2) \).
03

Calculate the Product Inside the Logarithm

Multiply the numbers inside the logarithm: \( 30 \times 2 = 60 \). Therefore, \( \log_{10} (30 \times 2) = \log_{10} 60 \).
04

Express as a Single Logarithm

Since you have simplified the expression to \( \log_{10} 60 \), this is already a single logarithm with a coefficient of 1. The process is complete.

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Key Concepts

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

Product Rule
The Product Rule is a fundamental property of logarithms that simplifies the addition of two logarithms with the same base. This rule states that adding two logs is equivalent to taking the logarithm of the product of their arguments. In mathematical terms, it can be expressed as:
  • \( \log_b M + \log_b N = \log_b (M \cdot N) \)
This rule is extremely useful because it helps transform a sum of two logarithms into a simpler single logarithmic expression.
For example, if you have \( \log_{10} 30 + \log_{10} 2 \), by applying the Product Rule, you multiply 30 and 2 to get 60. Thus, the expression becomes \( \log_{10} 60 \).
Using the Product Rule means fewer terms and easier calculations for complex logarithmic equations.
Logarithm Base
Understanding the base of a logarithm is crucial when you work with logarithmic expressions. The base determines how many times you need to multiply the base to obtain the number you are taking the log of. Mathematically, it can be defined as follows:
  • \( \log_b x \) answers the question, "What power do we need to raise \( b \) to, to obtain \( x \)?"
Common bases include 10 and \( e \) (the natural logarithm), but logs can have any positive value as a base. In the example, \( \log_{10} 30 \) and \( \log_{10} 2 \), the base 10 is often used in scientific calculations and is known as the common logarithm.
When combining logarithms, it is important to ensure that the bases of the logarithms are the same. Otherwise, the properties of logarithms, including the Product Rule, cannot be directly applied.
Combining Logarithms
Combining logarithms is a process of simplifying multiple logarithmic terms into a single term. To do this effectively, similar bases and certain properties, such as the Product Rule, must be utilized. Here's how you can approach combining logarithms:
  • Ensure the logarithms have the same base.
  • Apply appropriate logarithmic rules, like the Product Rule or Quotient Rule.
  • Simplify the resulting expression.
In the given exercise, combining \( \log_{10} 30 \) and \( \log_{10} 2 \) involves using the Product Rule because they have the same base of 10.
This simplification leads to a single logarithm \( \log_{10} 60 \). Combining logarithms effectively reduces complexity and makes calculation more straightforward, especially in algebraic or calculus-related problems.

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Most popular questions from this chapter

A function \(f\) with domain \((1, \infty)\) is defined by the equation \(f(x)=\log _{x} 2\) (a) Find a value for \(x\) such that \(f(x)=2\) (b) Is the number that you found in part (a) a fixed point of the function \(f ?\)

The age of some rocks can be estimated by measuring the ratio of the amounts of certain chemical elements within the rock. The method known as the rubidium-strontium method will be discussed here. This method has been used in dating the moon rocks brought back on the Apollo missions. Rubidium-87 is a radioactive substance with a half-life of \(4.7 \times 10^{10}\) years. Rubidium- 87 decays into the substance strontium- \(87,\) which is stable (nonradioactive). We are going to derive the following formula for the age of a rock: $$T=\frac{\ln \left[\left(\mathcal{N}_{s} / \mathcal{N}_{r}\right)+1\right]}{-k}$$ where \(T\) is the age of the rock, \(k\) is the decay constant for rubidium-87, \(\mathcal{N}_{s}\) is the number of atoms of strontium-87 now present in the rock, and \(\mathcal{N},\) is the number of atoms of rubidium-87 now present in the rock. (a) Assume that initially, when the rock was formed, there were \(\mathcal{N}_{0}\) atoms of rubidium-87 and none of strontium-87. Then, as time goes by, some of the rubidium atoms decay into strontium atoms, but the to tal number of atoms must still be \(\mathcal{N}_{0} .\) Thus, after \(T\) years, we have \(\mathcal{N}_{0}=\mathcal{N}_{r}+\mathcal{N}_{s}\) or, equivalently, $$ \mathcal{N}_{s}=\mathcal{N}_{0}-\mathcal{N}_{r}$$However, according to the law of exponential decay for the rubidium-87, we must have \(\mathcal{N}_{r}=\mathcal{N}_{0} e^{k T} .\) Solve this equation for \(\mathcal{N}_{0}\) and then use the result to eliminate \(\mathcal{N}_{0}\) from equation \((1) .\) Show that the result can be written $$\mathcal{N}_{s}=\mathcal{N}_{r} e^{-k T}-\mathcal{N}_{r}$$ (b) Solve equation (2) for \(T\) to obtain the formula given at the beginning of this exercise.

Simplify each expression. (a) \(\ln e\) (b) \(\ln e^{-2}\) (c) \((\ln e)^{-2}\)

Simplify each expression. (a) \(\ln e^{4}\) (b) \(\ln (1 / e)\) (c) \(\ln \sqrt{e}\)

(a) Use a graphing utility to estimate the root(s) of the equation to the nearest one-tenth (as in Example 6). (b) Solve the given equation algebraically by first rewriting it in logarithmic form. Give two forms for each answer: an exact expression and a calculator approximation rounded to three decimal places. Check to see that each result is consistent with the graphical estimate obtained in part (a). $$e^{t-1}=16$$

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