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Evaluate the logarithm. Round your result to three decimal places.\(\log _{20} 1575\)

Short Answer

Expert verified
The value of \( \log _{20} 1575 \) rounded to three decimal places is approximately 2.656.

Step by step solution

01

Understanding Logarithms

The logarithm of a number is the exponent to which another fixed number, the base, must be raised to produce that number. In this case, \( \log _{20} 1575 \) means finding the exponent that 20 must be raised to get 1575.
02

Calculating Logarithm

If we have a calculator that supports logarithms of any base, enter \( \log _{20} 1575 \) directly. If not, we can use the change of base formula - \( \log_b a = \frac{\log a}{\log b} \). Applying this to the current problem, we get that \( \log _{20} 1575 = \frac{\log 1575}{\log 20} \). Solving this with standard log (base 10 or e), the log values of 1575 and 20 can be found using a scientific calculator.
03

Rounding the Result

Once the result is obtained, it must be rounded to three decimal places as per the question.

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Key Concepts

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

Change of Base Formula
The change of base formula is a useful tool in mathematics when dealing with logarithms. Often, calculators only handle logarithms with certain bases, like base 10 or the natural base, e. This is where the change of base formula becomes especially helpful. It lets us rewrite a logarithm in terms of logs with a base that our calculator can handle.
In simple terms, if you have a logarithm \(\log_b a\), you can change its base by using the formula: \(\log_b a = \frac{\log_k a}{\log_k b}\). Here, \(k\) is any base that works with your calculator. Common choices are 10 (common logarithm) or \(e\) (natural logarithm).
Here's how you use it step-by-step:
  • Choose a new base \(k\) for the calculation.
  • Calculate \(\log_k a\).
  • Calculate \(\log_k b\).
  • Divide \(\log_k a\) by \(\log_k b\) to get your answer.
By applying this technique, you can find the value of logarithms that your calculator can't handle directly.
Exponentiation
Exponentiation is a mathematical operation involving two numbers: the base and the exponent. In exponentiation, the base is raised to the power of the exponent, which tells you how many times to multiply the base by itself. For example, if you have \(2^3\), it means multiplying 2 by itself three times: \(2 \times 2 \times 2 = 8\).
Logarithms are closely related to exponentiation. In fact, a logarithm is an exponentiation problem in reverse. The statement \(\log_b x = y\) means that \(b^y = x\). So when solving logarithmic problems, you're essentially figuring out which exponent turns one number into another.
Understanding this relationship can make it easier to grasp logarithms and the need for formulas like the change of base formula, since it allows us to solve these "reverse" operations even when we're limited by technology like calculators.
Scientific Calculator
A scientific calculator can help you solve mathematical problems that involve logarithms, exponentiation, and other operations. These calculators have functions that let you easily compute both base 10 and natural logarithms, represented by the keys "log" and "ln", respectively.
For logarithm calculations in different bases, you'd use the change of base formula to compute with your calculator's available logarithm functions. By converting to base 10 or base \(e\), you can get your logarithmic results, just like in the original exercise where \(\log_{20} 1575\) was calculated using \(\log_{10}\).
Besides logarithms, scientific calculators also handle exponentiation effortlessly with the "^" button or "exp" key, depending on the model.
  • To calculate exponentiation, simply enter the base number.
  • Press the "^" or "exp" key.
  • Follow with the exponent number, then press "equals" to get the result.
Ultimately, a scientific calculator is a powerful tool that simplifies many complex calculations, saving both time and ensuring accuracy. They're essential for students tackling higher math problems like those involving logarithms.

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