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Chapter 13: Problem 22

Find the natural logarithms of the given numbers. $$\sqrt{0.000060808}$$

Short Answer

Expert verified
The natural logarithm of \( \sqrt{0.000060808} \) is \(-4.853\).

Step by step solution

01

Simplify the Expression

First, simplify the expression \( \sqrt{0.000060808} \). This requires finding the square root of the number. To do this, we can express the number in scientific notation. \( 0.000060808 = 6.0808 \times 10^{-5} \).
02

Calculate the Square Root

Now, calculate the square root of the expression in scientific notation. The square root of \( 6.0808 \times 10^{-5} \) can be separated as \( \sqrt{6.0808} \times \sqrt{10^{-5}} \). Approximating \( \sqrt{6.0808} \approx 2.465 \) and noting \( \sqrt{10^{-5}} = 10^{-5/2} = 10^{-2.5} \), thus it becomes \( 2.465 \times 10^{-2.5} \).
03

Calculate the Natural Logarithm

The next step is calculating the natural logarithm of \( 2.465 \times 10^{-2.5} \). Using the logarithm property \( \ln(ab) = \ln(a) + \ln(b) \), we have \( \ln(2.465) + \ln(10^{-2.5}) \). Calculate \( \ln(2.465) \approx 0.902 \) and \( \ln(10^{-2.5}) = -2.5 \times \ln(10) \approx -2.5 \times 2.302 = -5.755 \).
04

Combine the Logarithms

Finally, add the logarithmic values together: \( 0.902 + (-5.755) = -4.853 \). Therefore, the natural logarithm of \( \sqrt{0.000060808} \) is \( -4.853 \).

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Key Concepts

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

Scientific Notation
Scientific notation is a way of expressing very large or very small numbers in a concise form. This method comes in handy, especially in scientific calculations where you deal with extreme values. A number in scientific notation is represented as the product of a number (usually between 1 and 10) and a power of 10.

  • For example, the number 0.000060808 is written in scientific notation as \( 6.0808 \times 10^{-5} \).
  • The part before the multiplication sign, \( 6.0808 \), is known as the coefficient.
  • The exponent \(-5\) signifies the decimal has been moved five places to the right to get the base value below 10.
This notation simplifies calculations as it breaks down complex numbers into more manageable forms. For example, in a multiplication or division involving numbers with powers of 10, you can just add or subtract the exponents.
Square Root Calculation
Calculating the square root is a common operation that can sometimes become complex, especially with very small or large numbers. For a number like 0.000060808, converting it to scientific notation helps.
  • First, express it as \( 6.0808 \times 10^{-5} \).
  • Then, find the square root of the coefficient \( 6.0808 \), which approximately equals \( 2.465 \).
  • The square root of \( 10^{-5} \) is \( 10^{-5/2} \) or \( 10^{-2.5} \).
This approach allows you to separately handle the components of the expression鈥攁n easier path than tackling the raw number. Once separated, you can directly compute typical square roots, multiplying the results back together for the final product.
Logarithm Properties
Logarithms transform multiplicative relationships into additive ones, which can make various mathematical problems easier to handle. In this particular problem, understanding the properties of logarithms is key.
  • The natural logarithm, denoted as \( \ln \), uses the constant \( e \) as its base, which is approximately 2.718.
  • One key property is \( \ln(ab) = \ln(a) + \ln(b) \). This allows you to split products into sums of logarithms.
  • For numbers represented in scientific notation, like \( 10^x \), you can use \( \ln(10^x) = x \ln(10) \).
In our exercise, using these properties allows us to find the natural logarithm of \( 2.465 \times 10^{-2.5} \) by splitting it into \( \ln(2.465) + \ln(10^{-2.5}) \). Each component is then easy to deal with: \( \ln(2.465) \approx 0.902 \) and \( \ln(10^{-2.5}) = -2.5 \times 2.302 = -5.755 \). Combining these results gives the final answer.

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Most popular questions from this chapter

Plot the indicated semilogarithmic graphs for the following application. In a particular electric circuit, called a low-pass filter, the input voltage \(V_{i}\) is across a resistor and a capacitor, and the output voltage \(V_{0}\) is across the capacitor (see Fig. 13.28 ). The voltage gain \(G\) (in \(d B\) ) is given by $$G=20 \log \frac{1}{\sqrt{1+(\omega T)^{2}}}$$ where \(\tan \phi=-\omega T\) Here, \(\phi\) is the phase angle of \(V_{0} / V_{i .}\) For values of \(\omega T\) of 0.01,0.1,0.3 \(1.0,3.0,10.0,30.0,\) and \(100,\) plot the indicated graphs. These graphs are called a Bode diagram for the circuit. Calculate values of \(\phi\) (as negative angles) for the given values of \(\omega T\) and plot a semilogarithmic graph of \(\phi\) vs. \(\omega T\)

Find \(f(\text { in } \mathrm{Hz})\) if \(\ln f=21.619,\) where \(f\) is the frequency of the microwaves in a microwave oven.

Use a calculator to display the indicated graphs. The graph of \(y=\log _{5} x\)

$$\text {Plot the indicated graphs.}$$ The electric power \(P\) (in \(\mathrm{W}\) ) in a certain battery as a function of the resistance \(R\) (in \(\Omega\) ) in the circuit is given by \(P=\frac{100 R}{(0.50+R)^{2}}\) Plot \(P\) as a function of \(R\) on semilog paper, using the logarithmic scale for \(R\) and values of \(R\) from \(0.01 \Omega\) to \(10 \Omega\)

Find the natural antilogarithms of the given logarithms. $$5.420$$
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