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Express the quantity in terms of base 10 logarithms. $$\log _{5} 10$$

Short Answer

Expert verified
\( \log_{5} 10 = \frac{1}{\log_{10} 5} \)

Step by step solution

01

Identify the Problem

The problem is to express \( \log_{5} 10 \) in terms of base 10 logarithms.
02

Use the Change of Base Formula

The change of base formula allows us to express a logarithm in any other base. It states that \( \log_{b} a = \frac{\log_{c} a}{\log_{c} b} \) for any positive number \( c \). Here, we'll use it with base 10 (common logarithm):\[ \log_{5} 10 = \frac{\log_{10} 10}{\log_{10} 5} \]
03

Simplify the Expression

Now, simplify the fraction. We know that \( \log_{10} 10 = 1 \) because 10 is the base of the log. Therefore:\[ \log_{5} 10 = \frac{1}{\log_{10} 5} \]

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Key Concepts

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

Base 10 Logarithms
Understanding base 10 logarithms, also known as common logarithms, can greatly simplify the examination of exponential relationships. A base 10 logarithm, denoted as \( \log_{10} \), tells us the power to which 10 must be raised to obtain a given number. This means, for example, that \( \log_{10} 100 = 2 \) because \( 10^2 = 100 \).
Base 10 is particularly useful because it aligns with our decimal number system. This makes calculations and interpretations more intuitive and straightforward.
Many scientific calculators and mathematical software default to base 10 logarithms, simplifying computations in science and engineering fields.
  • It's commonly used for measuring the intensity of earthquakes (Richter scale) and the loudness of sound (decibels).
  • The notation is simple, often written simply as \( \log \) without a base, assuming base 10.
Understanding this common base sets the stage for expressing other logarithms, such as \( \log_{5} 10 \), in terms of base 10.
Logarithmic Expressions
Logarithmic expressions often need to be manipulated for easier computation or application, especially when working with bases other than 10. This manipulation often involves the Change of Base Formula. This formula allows us to rewrite any logarithm in terms of another base, which can be especially helpful when using a standard calculator that only supports base 10.
According to the Change of Base Formula, \( \log_b a = \frac{\log_c a}{\log_c b} \). This means if you want to rewrite \( \log_{5} 10 \) using base 10, you can express it as \( \frac{\log_{10} 10}{\log_{10} 5} \). This method utilizes base 10 logarithms to simplify complex calculations.
  • It helps in converting non-standard logarithms into common or natural logarithms.
  • Understanding this formula aids in diverse fields like computer science where binary logs need conversion.
  • This conversion is crucial in problem-solving where specific bases are not easily calculated.
Using these conversions allows for greater flexibility in mathematical problem solving.
Simplifying Logarithms
The final yet crucial step in working with logarithms is simplifying. Simplification helps expose relationships and aids in further calculations. When expressing \( \log_{5} 10 \) in terms of base 10 logarithms, you reach the expression \( \frac{1}{\log_{10} 5} \).
Here's how that simplification works: since \( \log_{10} 10 = 1 \), you can simplify \( \frac{\log_{10} 10}{\log_{10} 5} \) to \( \frac{1}{\log_{10} 5} \).
This process reduces complexity:
  • It simplifies calculations by transforming an unfamiliar expression into an easily manageable form.
  • Understanding simplifications like \( \log_{10} 10 = 1 \) can enhance problem-solving speed.
  • This simplification is crucial in real-world applications like calculating interest rates or population growth, where base 10 logarithms are prevalent.
Simplifying logarithms is not just a mathematical exercise; it’s a tool for translating theory into practical results in everyday life.

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Most popular questions from this chapter

Estimate a value for \(x\) such that \(\log _{2} x=100 .\) Use the approximation \(10^{3}=2^{10}\) to express your answer as a power of 10 Answer: \(10^{30}\)

If \(a^{2}+b^{2}=7 a b,\) where \(a\) and \(b\) are positive, show that $$\log \left[\frac{1}{3}(a+b)\right]=\frac{1}{2}(\log a+\log b)$$ no matter which base is used for the logarithms (but it understood that the same base is used throughout).

The following extract is from an article by Kim Murphy that appeared in the Los Angeles Times on September 14 1994 CAIRO-Over a chorus of reservations from Latin America and Islamic countries still troubled about abortion and family issues, nearly 180 nations adopted a wide-ranging plan Tuesday on global population, the first in history to obtain partial endorsement from the Vatican. The plan, approved on the final day of the U.N. population conference here, for the first time tries to limit the growth of the world's population by preventing it from exceeding 7.2 billion people over the next two decades. (a) In 1995 the world population was 5.7 billion, with a relative growth rate of \(1.6 \% /\) year. Assuming continued exponential growth at this rate, make a projection for the world population in the year \(2020 .\) Round off the answer to one decimal place. How does your answer compare to the target value of 7.2 billion mentioned in the article? (b) As in part (a), assume that in 1995 the world population was 5.7 billion. Determine a value for the growth constant \(k\) so that exponential growth throughout the years \(1995-2020\) leads to a world population of 7.2 billion in the year 2020 .

Solve each equation and solve for \(x\) in terms of the other letters. $$T=T_{1}+\left(T_{0}-T_{1}\right) e^{-k x}$$

This exercise demonstrates the very slow growth of the natural logarithm function \(y=\ln x .\) We consider the following question: How large must \(x\) be before the graph of \(y=\ln x\) reaches a height of \(10 ?\) (a) Graph the function \(y=\ln x\) using a viewing rectangle that extends from 0 to 10 in the \(x\) -direction and 0 to 12 in the \(y\) -direction. Note how slowly the graph rises. Use the graphing utility to estimate the height of the curve (the \(y\) -coordinate) when \(x=10\) (b) since we are trying to see when the graph of \(y=\ln x\) reaches a height of 10 , add the horizontal line \(y=10\) to your picture. Next, adjust the viewing rectangle so that \(x\) extends from 0 to \(100 .\) Now use the graphing utility to estimate the height of the curve when \(x=100 .\) [As both the picture and the \(y\) -coordinate indicate, we're still not even halfway to \(10 .\) Go on to part (c).] (c) Change the vicwing rectangle so that \(x\) extends to \(1000,\) then estimate the \(y\) -coordinate corresponding to \(x=1000 .\) (You'll find that the height of the curve is almost \(7 .\) We're getting closer.) (d) Repeat part (c) with \(x\) extending to \(10,000 .\) (You'll find that the height of the curve is over \(9 .\) We're almost there. \()\) (e) The last step: Change the viewing rectangle so that \(x\) extends to \(100,000,\) then use the graphing utility to estimate the \(x\) -value for which \(\ln x=10 .\) As a check on your estimate, rewrite the equation \(\ln x=10\) in exponential form, and evaluate the expression that you obtain for \(x\)

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