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Chapter 6: Problem 1

Find these logarithms (or exponents): (a) \(\log _{2} 32\) (b) \(\log _{2}(1 / 32)\) (c) \(\log _{32}(1 / 32)\) (d) \(\log _{32} 2\) (e) \(\log _{10}(10 \sqrt{10})\) (I) \(\log _{2}\left(\log _{2} 16\right)\)ms.

Short Answer

Expert verified
(a) 5, (b) -5, (c) -1, (d) 1/5, (e) 1.5, (I) 2.

Step by step solution

01

Evaluate \( \log_{2} 32 \)

To find \( \log_{2} 32 \), we need to determine the power to which 2 must be raised to get 32. Since \( 2^5 = 32 \), we have \( \log_{2} 32 = 5 \).
02

Evaluate \( \log_{2}(1/32) \)

To find \( \log_{2}(1/32) \), we recognize that \( 1/32 = 2^{-5} \). The logarithm is the power, so \( \log_{2}(1/32) = -5 \).
03

Evaluate \( \log_{32}(1/32) \)

To find \( \log_{32}(1/32) \), notice that \( 32^1 = 32 \) and \( 32^{-1} = 1/32 \). Therefore, \( \log_{32}(1/32) = -1 \).
04

Evaluate \( \log_{32} 2 \)

To find \( \log_{32} 2 \), we use the change of base formula: \( \log_{32} 2 = \frac{\log_{2} 2}{\log_{2} 32} \). Since \( \log_{2} 2 = 1 \) and \( \log_{2} 32 = 5 \), we have \( \log_{32} 2 = \frac{1}{5} \).
05

Evaluate \( \log_{10}(10\sqrt{10}) \)

Break it into a product: \( 10\sqrt{10} = 10 \times 10^{0.5} = 10^{1+0.5} = 10^{1.5} \). Thus \( \log_{10}(10\sqrt{10}) = \log_{10}(10^{1.5}) = 1.5 \).
06

Evaluate \( \log_{2} \left( \log_{2} 16 \right) \)

First find \( \log_{2} 16 \). Since \( 2^4 = 16 \), \( \log_{2} 16 = 4 \). Now find \( \log_{2} 4 \); since \( 2^2 = 4 \), we have \( \log_{2} (\log_{2} 16) = 2 \).

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Key Concepts

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

Exponents
An exponent is a way to express repeated multiplication of the same number. For example, writing \(2^5\) means that the number 2 is multiplied by itself 5 times, i.e., \(2 \times 2 \times 2 \times 2 \times 2\). This equals 32.Exponents are a fundamental concept in mathematics, as they provide a compact way to represent large numbers and conduct calculations more easily.
  • Base: The number that is multiplied.
  • Exponent: The power or number of times the base is multiplied.
In logarithmic equations, exponents play a key role as logs essentially ask "how many times do we multiply the base to get a particular number?" This is why understanding the relationship between exponentiation and logarithms is crucial for solving log equations.
Change of Base Formula
The change of base formula is a useful method when solving logarithms with bases other than 10 or \(e\), which are not typically calculated with a standard calculator.The formula is as follows:\[\log_b a = \frac{\log_c a}{\log_c b}\]This allows us to change the base of a logarithm from \(b\) to any base \(c\) (usually 10 or \(e\), for calculator convenience).For example, to find \(\log_{32} 2\), you can use this formula:
  • Let the new base be 2, then \(\log_{32} 2 = \frac{\log_2 2}{\log_2 32}\).
  • Compute these values: \(\log_2 2 = 1\) and \(\log_2 32 = 5\).
  • Thus, \(\log_{32} 2 = \frac{1}{5}\).
This simplifies computation and helps in solving many logarithmic equations efficiently.
Properties of Logarithms
Logarithms have several key properties that make them handy tools in simplifying and solving mathematical equations.
  • Product Property: \(\log_b (xy) = \log_b x + \log_b y\)
  • Quotient Property: \(\log_b \left(\frac{x}{y}\right) = \log_b x - \log_b y\)
  • Power Property: \(\log_b (x^n) = n \cdot \log_b x\)
These properties allow for transforming complex log expressions into simpler forms, which makes them easier to evaluate. For instance, finding \(\log_{10}(10\sqrt{10})\) was simplified using the power rule: it was rewritten as \(\log_{10}(10^{1.5})\), yielding a straightforward result of 1.5.
Logarithmic Equations
Logarithmic equations involve finding the unknown value in an equation set in a logarithmic form. Here, understanding and applying logs and their properties help solve these equations. Here's a typical approach:
  • Identify the form of the logarithmic equation.
  • Use known properties to simplify the equation, if possible.
  • Solve for the unknown by rewriting the equation into an exponential form if needed.
For instance, to solve \(\log_{2} \left(\log_{2} 16\right)\), first compute \(\log_{2} 16\), which is 4, and then move to \(\log_{2} 4\) which solves to 2. This layered approach helps in breaking down and unraveling complex log problems step by step.

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Most popular questions from this chapter

Solve with \(y_{0}=0\) and graph the solution. $$ \frac{d y}{d t}=y+1 $$

A solitary water wave has a shape satisfying the \(K d V\) equation \(y^{\prime \prime}=y^{\prime}-6 y y^{\prime}\) (a) Integrate once to find \(y^{\prime \prime} .\) Multiply the answer by \(y^{\prime}\). (b) Integrate again to find \(y^{\prime}\) (all constants of integration are zero). (c) Show that \(y=\frac{1}{2} \operatorname{sech}^{2}(x / 2)\) gives the shape of the "soliton."

A falling body with friction equal to velocity squared obeys \(d v / d t=g-v^{2}\) (a) Show that \(v(t)=\sqrt{g}\) tanh \(\sqrt{g t \text { satisfies the equation. }}\) (b) Derive this \(v\) yourself, by integrating \(d v /\left(g-v^{2}\right)=d t\) (c) Integrate \(v(t)\) to find the distance \(f(t)\)

Solve with \(y_{0}=0\) and graph the solution. $$ \frac{d y}{d t}=-y+1 $$

Prove that \(\ln x \leqslant 2(\sqrt{x}-1)\) for \(x>1 .\) Compare the integrats of \(1 / t\) and \(1 / \sqrt{t},\) from 1 to \(x\)
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