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Exer. \(55-56:\) Approximate \(x\) to three significant figures. (a) \(\log x=3.6274\) (b) \(\log x=0.9469\) (c) \(\log x=-1.6\) (d) \(\ln x=2.3\) (e) \(\ln x=0.05\) (f) \(\ln x=-1.6\)

Short Answer

Expert verified
(a) 4241, (b) 8.90, (c) 0.0251, (d) 9.97, (e) 1.05, (f) 0.202.

Step by step solution

01

Understanding the Exercise

We need to find the value of \(x\) for each part of this exercise, given a logarithmic or natural logarithmic equation, and round to three significant figures. We will first convert the log equations to exponential form.
02

Solve for x in \(\log x = 3.6274\)

Given \(\log x = 3.6274\), this is a base 10 logarithm. Convert it to exponential form: \[ 10^{3.6274} = x \]Calculate \(x\): \[ x = 10^{3.6274} \approx 4.241 \times 10^{3} \] The value of \(x\) is approximately 4241.
03

Solve for x in \(\log x = 0.9469\)

Given \(\log x = 0.9469\), convert it to exponential form: \[ 10^{0.9469} = x \]Calculate \(x\): \[ x = 10^{0.9469} \approx 8.90 \] The value of \(x\) is approximately 8.90.
04

Solve for x in \(\log x = -1.6\)

Given \(\log x = -1.6\), convert it to exponential form: \[ 10^{-1.6} = x \]Calculate \(x\): \[ x = 10^{-1.6} \approx 0.0251 \] The value of \(x\) is approximately 0.0251.
05

Solve for x in \(\ln x = 2.3\)

Given \(\ln x = 2.3\), convert it to exponential form using base \(e\): \[ e^{2.3} = x \]Calculate \(x\): \[ x = e^{2.3} \approx 9.97 \] The value of \(x\) is approximately 9.97.
06

Solve for x in \(\ln x = 0.05\)

Given \(\ln x = 0.05\), convert it to exponential form using base \(e\): \[ e^{0.05} = x \]Calculate \(x\): \[ x = e^{0.05} \approx 1.05 \] The value of \(x\) is approximately 1.05.
07

Solve for x in \(\ln x = -1.6\)

Given \(\ln x = -1.6\), convert it to exponential form using base \(e\): \[ e^{-1.6} = x \]Calculate \(x\): \[ x = e^{-1.6} \approx 0.202 \] The value of \(x\) is approximately 0.202.

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Key Concepts

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

Significant Figures
Significant figures are crucial in mathematics and measurements. They tell us about the precision of our calculated results. Understanding significant figures can seem complicated, but it's easier with practice. When we say 'three significant figures,' we mean the first three non-zero digits of a number. For example, in the number 4241, the significant figures are 4, 2, and 1.
It's important to note that zeros can sometimes count and sometimes do not.
  • Leading zeros (zeros in front of the first non-zero digit) are not significant. For instance, 0.0251 has three significant figures: 2, 5, and 1.
  • Captive zeros (zeros between other non-zero digits) are significant. In the number 1002, all four digits are considered significant figures.
  • Trailing zeros, which are zeros at the end of a number, count as significant only if there is a decimal point. For example, 45.00 has four significant figures. Without the decimal, 4500 would only have two significant figures if not otherwise specified.
In calculations, maintaining the correct number of significant figures ensures precision without implying misleading accuracy.
Base 10 Logarithms
Base 10 logarithms, also known as common logarithms, use 10 as the base. In mathematics, logarithms help us solve for values in exponential equations. For example, if you see a problem like \(\log x = 3.6274\), it means that 10 raised to the power of something equals \(x\).
To reverse a logarithm and find \(x\), we convert the equation to its exponential form:
  • Given \(\log x = a\), we write it as \(10^a = x\).
  • Now, \(x\) can be calculated as \(10^{3.6274} \approx 4241\).
This concept is widely used in science and engineering to simplify multiplication and division into addition and subtraction, thanks to properties of logarithms.
Typically, log tables or calculators are used for these conversions because calculating them manually could be tedious. Knowing how to interpret base 10 logarithms is crucial for mastering exponential-related problems.
Natural Logarithms
Natural logarithms use the mathematical constant \(e\) as their base, roughly equal to 2.71828. Most often represented as \(\ln\), they help in modeling continuous growth or decay, commonly appearing in calculus and complex science problems.
For instance, in equations like \(\ln x = 2.3\), we convert it to exponential form using base \(e\):
  • The equation becomes \(e^{2.3} = x\).
  • This translates into \(x \approx 9.97\).
It's the inverse operation of finding a power for a given exponential function. Natural logarithms are instrumental in calculating more complex equations efficiently.
Using natural logarithms, often done with calculators because of \(e\)'s non-terminating nature, makes it easier to deal with exponential modifications like growth rates in economy or decay rates in physical processes.

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Most popular questions from this chapter

Exer. \(51-54\) : Chemists use a number denoted by \(p H\) to describe quantitatively the acidity or basicity of solutions. By definition, \(\mathbf{p H}=-\log \left[\mathbf{H}^{+}\right],\) where \(\left[\mathbf{H}^{+}\right]\) is the hydrogen ion concentration in moles per liter. A solution is considered basic if \(\left[\mathrm{H}^{+}\right]<10^{-7}\) or acidic if \(\left[\mathrm{H}^{+}\right]>10^{-7} .\) Find the corresponding inequalities involving pH.

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