If log 10 2 = 0.3010 and log 10 3 = 0.4771, then the value of log 100 (0.72) is equal to
Calculating Logarithms: Understanding the Problem
The question asks us to find the value of \(\log_{100}(0.72)\) using the given values of \(\log_{10} 2\) and \(\log_{10} 3\). This involves using logarithm properties, specifically the change of base formula, to convert the given logarithm to base 10.
We are given:
- \(\log_{10} 2 = 0.3010\)
- \(\log_{10} 3 = 0.4771\)
We need to calculate \(\log_{100}(0.72)\).
Steps to Solve Logarithm Calculation
To solve this problem, we will follow these steps:
- Use the change of base formula to convert \(\log_{100}(0.72)\) to base 10.
- Simplify the argument of the logarithm (0.72) into terms involving 2 and 3.
- Apply logarithm properties to express \(\log_{10}(0.72)\) in terms of \(\log_{10} 2\) and \(\log_{10} 3\).
- Substitute the given values of \(\log_{10} 2\) and \(\log_{10} 3\) into the expression.
- Perform the necessary arithmetic calculations.
- Express the final result in the characteristic-mantissa form if needed, matching the options provided.
Detailed Calculation of \(\log_{100}(0.72)\)
First, let's apply the change of base formula. The formula is \(\log_b a = \frac{\log_c a}{\log_c b}\). We will change the base from 100 to 10:
\(\log_{100}(0.72) = \frac{\log_{10}(0.72)}{\log_{10}(100)}\)
Now, let's evaluate the denominator \(\log_{10}(100)\):
\(\log_{10}(100) = \log_{10}(10^2) = 2\) (since \(\log_b b^x = x\))
Next, let's evaluate the numerator \(\log_{10}(0.72)\). We can write 0.72 as a fraction:
\(0.72 = \frac{72}{100}\)
So, \(\log_{10}(0.72) = \log_{10}\left(\frac{72}{100}\right)\). Using the logarithm property \(\log \frac{a}{b} = \log a - \log b\):
\(\log_{10}\left(\frac{72}{100}\right) = \log_{10}(72) - \log_{10}(100)\)
We already know \(\log_{10}(100) = 2\). Now we need to evaluate \(\log_{10}(72)\). We can factorize 72 into prime numbers involving 2 and 3:
\(72 = 8 \times 9 = 2^3 \times 3^2\)
Using the logarithm property \(\log(ab) = \log a + \log b\) and \(\log(a^p) = p \log a\):
\(\log_{10}(72) = \log_{10}(2^3 \times 3^2) = \log_{10}(2^3) + \log_{10}(3^2) = 3 \log_{10} 2 + 2 \log_{10} 3\)
Now substitute the given values \(\log_{10} 2 = 0.3010\) and \(\log_{10} 3 = 0.4771\):
\(\log_{10}(72) = 3(0.3010) + 2(0.4771)\)
| Term | Calculation | Value |
|---|---|---|
| \(3 \log_{10} 2\) | \(3 \times 0.3010\) | 0.9030 |
| \(2 \log_{10} 3\) | \(2 \times 0.4771\) | 0.9542 |
| \(\log_{10}(72)\) | \(0.9030 + 0.9542\) | 1.8572 |
So, \(\log_{10}(72) = 1.8572\).
Now, substitute this back into the expression for \(\log_{10}(0.72)\):
\(\log_{10}(0.72) = \log_{10}(72) - \log_{10}(100) = 1.8572 - 2 = -0.1428\)
Finally, substitute the values for the numerator and denominator into the change of base formula expression:
\(\log_{100}(0.72) = \frac{\log_{10}(0.72)}{\log_{10}(100)} = \frac{-0.1428}{2} = -0.0714\)
Interpreting the Result and Matching Options
The calculated value is \(-0.0714\). The options are given in the format \(\bar{X}.YYYY\). This format represents a negative logarithm where the integer part (characteristic) is negative, and the decimal part (mantissa) is positive.
To convert \(-0.0714\) into this format, we add and subtract 1:
\(-0.0714 = -1 + 1 - 0.0714 = -1 + (1 - 0.0714) = -1 + 0.9286\)
In characteristic-mantissa form, \(-1 + 0.9286\) is written as \(\bar{1}.9286\). The characteristic is \(\bar{1}\) (which means -1) and the mantissa is \(0.9286\).
Comparing this result with the given options, we find that it matches option 2.
Revision Table: Key Logarithm Concepts
| Concept | Formula/Property | Explanation |
|---|---|---|
| Change of Base | \(\log_b a = \frac{\log_c a}{\log_c b}\) | Allows conversion between logarithm bases. |
| Product Rule | \(\log_b (xy) = \log_b x + \log_b y\) | Logarithm of a product is the sum of logarithms. |
| Quotient Rule | \(\log_b \left(\frac{x}{y}\right) = \log_b x - \log_b y\) | Logarithm of a quotient is the difference of logarithms. |
| Power Rule | \(\log_b (x^p) = p \log_b x\) | Logarithm of a power is the exponent times the logarithm. |
| Logarithm of Base | \(\log_b b = 1\) | The logarithm of the base itself is 1. |
| Logarithm of 1 | \(\log_b 1 = 0\) | The logarithm of 1 is always 0. |
| Characteristic & Mantissa | \(\log_{10} N = \text{Characteristic} + \text{Mantissa}\) | Characteristic is the integer part, mantissa is the non-negative decimal part (\(0 \le \text{Mantissa} < 1\)). \(\bar{X}.YYYY\) means Characteristic is -X. |
Additional Information on Logarithms
Logarithms are the inverse operation to exponentiation. The expression \(\log_b a = c\) means that \(b^c = a\). Base 10 logarithms, also known as common logarithms, are widely used in science and engineering.
The characteristic and mantissa form is a standard way to represent logarithms, especially before the widespread use of calculators, as logarithm tables typically provided only the mantissa, and the characteristic was determined by the position of the decimal point in the original number.
When the number is between 0 and 1, its base 10 logarithm is negative. For example, \(\log_{10}(0.1) = -1\), \(\log_{10}(0.01) = -2\). The characteristic for a number less than 1 is negative and is determined by the number of zeros immediately after the decimal point plus one, with a negative sign. For example, for 0.072, the decimal point is after one zero, so the characteristic is -2. However, when calculating the logarithm value itself, as we did, we get the actual negative number. Converting this to the \(\bar{X}.YYYY\) form involves making the decimal part positive.
In our calculation, \(\log_{10}(0.72) = -0.1428\). For a number like 0.72 (between 0.1 and 1), the characteristic of its base 10 log is -1. We see this in the form \(\bar{1}.9286\), where the characteristic is indeed -1.
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