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Exer. 9-10: Evaluate using the change of base formula (without a calculator). $$\frac{\log _{5} 16}{\log _{5} 4}$$

Short Answer

Expert verified
The expression evaluates to 2.

Step by step solution

01

Understand the Change of Base Formula

The change of base formula allows you to rewrite logarithms in terms of logarithms with a different base. It is given by: \( \log_b a = \frac{\log_c a}{\log_c b} \). This will be useful as we need to use common bases for simpler calculations.
02

Apply the Change of Base Formula

Use the change of base formula to rewrite both the numerator and the denominator in terms of base 10 (or any common base). This gives: \( \frac{\log_{10} 16}{\log_{10} 5} \) for the numerator and \( \frac{\log_{10} 4}{\log_{10} 5} \) for the denominator.
03

Simplify the Expression

Now, substitute the expressions from Step 2 into the original fraction: \( \frac{\frac{\log_{10} 16}{\log_{10} 5}}{\frac{\log_{10} 4}{\log_{10} 5}} \). This simplifies to \( \frac{\log_{10} 16}{\log_{10} 4} \) as the \( \log_{10} 5 \) terms cancel out.
04

Further Simplify the Logarithms

Recognize that \( \log_{10} 16 = \log_{10} (2^4) = 4 \log_{10} 2 \) and \( \log_{10} 4 = \log_{10} (2^2) = 2 \log_{10} 2 \). Substitute these back into the expression to get \( \frac{4 \log_{10} 2}{2 \log_{10} 2} \).
05

Final Calculation

Cancel out the \( \log_{10} 2 \) terms in the fraction: \( \frac{4}{2} = 2 \).
06

Conclusion

After simplifying the expression, we find that the original expression evaluates to 2.

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Key Concepts

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

Logarithmic Functions
Logarithmic functions are an essential part of algebra and calculus. A logarithm is essentially the inverse operation of exponentiation. When you see something like \( \log_b a \), it is asking the question: "To what power must \( b \) be raised, to yield \( a \)?"
Understanding logarithms is crucial because they allow us to solve equations where the variable is in an exponent. This makes them particularly useful for dealing with exponential growth or decay problems, like population growth or radioactive decay.
  • Standard Notation: \( \log_b a \) is used to denote the logarithm of \( a \) with base \( b \).
  • Inverse Relationship: If \( b^x = a \), then \( \log_b a = x \).
Getting comfortable with logarithms requires practice, especially when dealing with different bases. Many problems will require you to manipulate them into a more workable form when solving complex equations.
Base Conversion
Base conversion with logarithms is a powerful technique that helps when working with unfamiliar or unwieldy bases. The Change of Base Formula is the tool that allows this conversion:
\[ \log_b a = \frac{\log_c a}{\log_c b} \]
This formula enables the conversion of a logarithm from one base to another. It's particularly handy when dealing with calculators that primarily support base 10 logarithms.
Here's how base conversion takes shape in mathematics:
  • You can express any logarithm in terms of common logarithms (base 10) or natural logarithms (base e).
  • The Change of Base Formula is often used to simplify calculations by using more familiar bases.
In practical terms, this formula means you can solve logarithmic equations without needing complex base support on your calculator. Instead, you convert everything to a common base and simplify the calculations from there.
Logarithm Simplification
Simplifying logarithms can make solving equations much clearer and more straightforward. Once logarithms are expressed in terms of a single base, they can often be simplified using basic properties of logarithms. For example, consider the expression \( \frac{\log_{10} 16}{\log_{10} 4} \).
We can simplify this by recognizing the properties of logarithms:
  • Power Rule: \( \log_b (a^c) = c \cdot \log_b a \). This means you can take exponents "out front" of the log as multiplication.
  • Quotient Rule: \( \frac{\log_b a}{\log_b c} = \log_b a - \log_b c \), simplifying division expressions to subtractions.
Applying these rules allows us to rewrite logarithmic terms for easier computation. In our example, recognizing that \( 16 = 2^4 \) and \( 4 = 2^2 \) allows the use of the power rule:
\[ \frac{4 \cdot \log_{10} 2}{2 \cdot \log_{10} 2} = 2 \]
The terms for \( \log_{10} 2 \) cancel out, resulting in a much simpler equation to solve. By mastering these simplification techniques, you can handle more complex logarithmic expressions with ease.

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Most popular questions from this chapter

Exer. 83-84: Approximate the function at the value of \(x\) to four decimal places. (a) \(f(x)=\ln (x+1)+e^{x}, \quad x=2\) (b) \(g(x)=\frac{(\log x)^{2}-\log x}{4}, \quad x=3.97\)

Air pressure The air pressure \(p(h)\) (in \(\mathrm{Ib} / \mathrm{in}^{2}\) ) at an altitude of \(h\) feet above sea level may be approximated by the formula \(p(h)=14.7 e^{-0.0000385 h}\). At approximately what altitude \(h\) is the air pressure (a) \(10 \mathrm{lb} / \mathrm{in}^{2} ?\) (b) one-half its value at sea level?

Sketch the graph of the equation. (a) Estimate \(y\) if \(x=40\). (b) Estimate \(x\) if \(y=2\). $$y=(1.085)^{x}$$

In 1840 , Britain experienced a bovine (cattle and oxen) epidemic called epizooty. The estimated number of new cases every 28 days is listed in the table. At the time, the London Daily made a dire prediction that the number of new cases would continue to increase indefinitely. William Farr correctly predicted when the number of new cases would peak. Of the two functions $$\begin{array}{l}f(t)=653(1.028)^{t} \\\g(t)=54,700 e^{-(t-200)^{2} / 7500}\end{array}$$ and one models the newspaper's prediction and the other models Farr's prediction, where \(t\) is in days with \(t=0\) corresponding to August \(12,1840\). $$\begin{array}{|c|c|}\hline \text { Date } & \text { New cases } \\\\\hline \text { Aug. 12 } & 506 \\\\\hline \text { Sept. 9 } & 1289 \\\\\hline \text { Oct. 7 } & 3487 \\\\\hline \text { Nov. 4 } & 9597 \\\\\hline \text { Dec. 2 } & 18,817 \\\\\hline \text { Dec. 30 } & 33,835 \\\\\hline \text { Jan. 27 } & 47,191 \\\\\hline\end{array}$$ (a) Graph each function, together with the data, in the viewing rectangle \([0,400,100]\) by \([0,60,000,10,000]\) (b) Determine which function better models Farr's prediction. (c) Determine the date on which the number of new cases peaked.

Exer. 83-84: Approximate the function at the value of \(x\) to four decimal places. (a) \(f(x)=\log \left(2 x^{2}+1\right)-10^{-x}, \quad x=1.95\) (b) \(g(x)=\frac{x-3.4}{\ln x+4}, \quad x=0.55\)

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