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Suppose that \(\log _{10} A=a, \log _{10} B=b\) and \(\log _{10} C=c .\) Express the following logarithms in terms of \(a, b,\) and \(c\) (a) \(\log _{10} A B^{2} C^{3}\) (b) \(\log _{10} 10 \sqrt{A}\) (c) \(\log _{10} \sqrt{10 A B C}\) (d) \(\log _{10}(10 A / \sqrt{B C})\)

Short Answer

Expert verified
(a) a + 2b + 3c; (b) 1 + a/2; (c) (1 + a + b + c)/2; (d) 1 + a - (b+c)/2.

Step by step solution

01

Analyzing Problem (a)

To find \( \log_{10}(AB^2C^3) \), we use the properties of logarithms which state that \( \log_{10}(XY) = \log_{10} X + \log_{10} Y \) and \( \log_{10}(X^n) = n\log_{10} X \). Start by expanding \( \log_{10}(AB^2C^3) \) using these properties.
02

Applying Logarithmic Properties to (a)

For \( \log_{10}(AB^2C^3) \), apply the properties: \[ \log_{10}(A) + \log_{10}(B^2) + \log_{10}(C^3) \]. Substitute the known logarithms: \[ a + 2b + 3c \].
03

Solution to (a)

After substituting the expressions, the result for (a) is \( \log_{10}(AB^2C^3) = a + 2b + 3c \).
04

Analyzing Problem (b)

For \( \log_{10}(10\sqrt{A}) \), decompose it as \( \log_{10}(10) + \log_{10}(A^{1/2}) \). Using properties, this becomes \( 1 \) (since \( \log_{10}(10) = 1 \)) plus \( \frac{1}{2} \log_{10}(A) \).
05

Solution to (b)

Substituting \( \log_{10}(A) = a \), we get \( \log_{10}(10\sqrt{A}) = 1 + \frac{a}{2} \).
06

Analyzing Problem (c)

For \( \log_{10}(\sqrt{10ABC}) \), recognize it as \( \log_{10}((10ABC)^{1/2}) \). Apply the power rule: \( \frac{1}{2}(\log_{10}(10) + \log_{10}(A) + \log_{10}(B) + \log_{10}(C)) \).
07

Solution to (c)

This becomes \( \frac{1}{2}(1 + a + b + c) = \frac{1 + a + b + c}{2} \).
08

Analyzing Problem (d)

Express \( \log_{10}(10A/\sqrt{BC}) \) as \( \log_{10}(10A) - \log_{10}(\sqrt{BC}) \). Simplifying, \( \log_{10}(10A) = \log_{10}(10) + \log_{10}(A) = 1 + a \), and \( \log_{10}(\sqrt{BC}) = \frac{1}{2}(\log_{10}(B) + \log_{10}(C)) = \frac{b+c}{2} \).
09

Solution to (d)

Substitute the expressions to find: \( \log_{10}(10A/\sqrt{BC}) = (1 + a) - \frac{b+c}{2} = 1 + a - \frac{b+c}{2} \).

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Key Concepts

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

Logarithm Properties
Logarithms are fascinating mathematical tools used to simplify complex equations, especially those involving exponential growth or decay. Understanding the properties of logarithms is vital to solving logarithmic expressions and equations.
These properties can make handling complex expressions such as
  • Product rule: \(\log_{10}(XY) = \log_{10}X + \log_{10}Y\)
  • Power rule: \(\log_{10}(X^n) = n \log_{10}X\)
  • Quotient rule: \(\log_{10}(\frac{X}{Y}) = \log_{10}X - \log_{10}Y\)
These allow us to break down expressions into simpler terms. For instance, multiplying elements inside a logarithm becomes an addition of their individual logs, raising a number to a power becomes a multiplication, and dividing becomes subtraction between logs.

In the given exercise, these properties simplify expressions like \(\log_{10}(AB^2C^3)\) into \(a + 2b + 3c\). Expanding each component separately using these rules ensures a simpler, more manageable expression which is crucial when dealing with logarithms in precalculus math.
Logarithmic Equations
Logarithmic equations involve expressions set equal to a logarithmic expression. Solving these requires a good grasp of logarithm properties. Every step utilizes the properties of logarithms to condense or expand parts of the expression.
For instance, consider an equation like \(\log_{10}(10\sqrt{A})\). You can break this into \(\log_{10}(10) + \log_{10}(A^{1/2})\), applying properties of logs.
  • The simplification \(\log_{10}(10) = 1\), since 10 is the base of our logarithm.
  • Moreover, using the power rule, \(\log_{10}(A^{1/2}) = \frac{1}{2} \log_{10}A\).
Next, substitute your known values to express it in terms of variables like \(a, b, \) and \(c\).

These strategies are essential in precalculus math as they build a foundation for understanding more complex topics in calculus. With practice, solving logarithmic equations becomes easier, making the seemingly complex notion very approachable.
Precalculus Math
Precalculus is a branch of mathematics that prepares students for understanding calculus. It revisits and fortifies foundational concepts with advanced topics like trigonometry and complex numbers, including significant focus on functions such as logarithms.
Logarithmic functions are particularly significant in precalculus due to their powerful nature in modeling exponentials, a concept recurrent in calculus. They prepare students to handle growth models, sound intensity, and pH levels, which can involve various logarithmic operations.
Within the scope of precalculus, you learn to interpret logarithmic expressions, manipulate them using the learned properties, and solve real-world problems that frame exponential relationships.
  • Grasping logarithmic rules enables precise calculations of exponential growth like in population models.
  • Moreover, understanding these functions helps ease into limits and derivatives by simplifying complex expressions.
Solidifying these concepts not only aids comprehension in calculus but also fosters great analytical skills required across different scientific fields.

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Most popular questions from this chapter

(a) Graph the two functions \(f(x)=(\ln x) /(\ln 3)\) and \(g(x)=\ln x-\ln 3 .\) (Use a viewing rectangle in which \(x\) extends from 0 to 10 and \(y\) extends from \(-5 \text { to } 5 .)\) Why aren't the two graphs identical? That is, doesn't one of the basic log identities say that \((\ln a) /(\ln b)=\ln a-\ln b ?\) (b) Your picture in part (a) indicates that $$\frac{\ln x}{\ln 3}>\ln x-\ln 3 \quad(0< x \leq 10)$$ Find a viewing rectangle in which \((\ln x) /(\ln 3) \leq \ln x-\ln 3.\) (c) Use the picture that you obtain in part (b) to estimate the value of \(x\) for which \((\ln x) /(\ln 3)=\ln x-\ln 3.\) (d) Solve the equation \((\ln x) /(\ln 3)=\ln x-\ln 3\) algebraically and use the result to check your estimate in part (c).

The Chernobyl nuclear explosion (in the former Soviet Union, on April 26,1986 ) released large amounts of radioactive substances into the atmosphere. These substances included cesium-137, iodine-131, and strontium-90. Although the radioactive material covered many countries, the actual amount and intensity of the fallout varied greatly from country to country, due to vagaries of the weather and the winds. One area that was particularly hard hit was Lapland, where heavy rainfall occurred just when the Chernobyl cloud was overhead. (a) Many of the pastures in Lapland were contaminated with cesium-137, a radioactive substance with a half- life of 33 years. If the amount of cesium- 137 was found to be ten times the normal level, how long would it take until the level returned to normal? Hint: Let \(\mathcal{N}_{0}\) be the amount that is ten times the normal level. Then you want to find the time when \(\mathcal{N}(t)=\mathcal{N}_{0} / 10\) (b) Follow part (a), but assume that the amount of cesium-137 was 100 times the normal level. Remark: Several days after the explosion, it was reported that the level of cesium- 137 in the air over Sweden was 10,000 times the normal level. Fortunately there was little or no rainfall.

In the text we showed that the relative growth rate for the function \(\mathcal{N}(t)=\mathcal{N}_{0} e^{k t}\) is constant for all time intervals of unit length, \([t, t+1] .\) Recall that we did this by computing the relative change \([\mathcal{N}(t+1)-\mathcal{N}(t)] / \mathcal{N}(t)\) and noting that the result was a constant, independent of \(t .\) (If you've completed the previous exercise, you've done this calculation for yourself.) Now consider a time interval of arbitrary length, \([t, t+d] .\) The relative change in the function \(\mathcal{N}(t)=\mathcal{N}_{0} e^{k t}\) over this time interval is \([\mathcal{N}(t+d)-\mathcal{N}(t)] / \mathcal{N}(t) .\) Show that this quantity is a constant, independent of \(t .\) (The expression that you obtain for the constant will contain \(e\) and \(d\), but not \(t\) As a check on your work, replace \(d\) by 1 in the expression you obtain and make sure the result is the same as that in the text where we worked with intervals of length \(d=1 .)\)

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.

Decide which of the following properties apply to each function. (More than one property may apply to a function.)A. The function is increasing for \(-\infty

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