Let XYZ be an equilateral triangle in which XY = 7 cm. If A denotes the area of the triangle, then what is the value of log 10 A4 ? (Given that log 10 1050 = 3.0212 and log 10 35 = 1.5441)
5.3070
Understanding the Equilateral Triangle Area and Logarithms Problem
The problem asks us to calculate the value of \(\log_{10} A^4\), where \(A\) represents the area of an equilateral triangle with a side length of 7 cm. We are also provided with two logarithmic values: \(\log_{10} 1050 = 3.0212\) and \(\log_{10} 35 = 1.5441\). These given values are key to solving the problem without needing a calculator for standard logarithm values like \(\log_{10} 2\), \(\log_{10} 3\), etc.
Calculating the Area of the Equilateral Triangle
The area \(A\) of an equilateral triangle with side length \(s\) is given by the formula:
$\(A = \frac{\sqrt{3}}{4} s^2$\)
In this problem, the side length \(s = 7\) cm. Substituting this value into the formula:
$\(A = \frac{\sqrt{3}}{4} (7 \, \text{cm})^2 = \frac{\sqrt{3}}{4} \times 49 \, \text{cm}^2 = \frac{49\sqrt{3}}{4} \, \text{cm}^2$\)
Evaluating the Logarithmic Expression \(\log_{10} A^4\)
We need to find the value of \(\log_{10} A^4\). Using the logarithm property \(\log b^c = c \log b\), we can rewrite the expression:
$\(\log_{10} A^4 = 4 \log_{10} A$\)
Substitute the calculated value of \(A\) into the expression:
$\(4 \log_{10} \left(\frac{49\sqrt{3}}{4}\right)$\)
Now, we need to evaluate \(\log_{10} \left(\frac{49\sqrt{3}}{4}\right)\) using the given logarithmic values. Let's try to express the argument \(\frac{49\sqrt{3}}{4}\) in a way that relates to 1050 or 35.
We know \(49 = 7^2\). Also, \(35 = 5 \times 7\). So, \(7 = \frac{35}{5}\).
Let's rewrite the area \(A\) using this relationship:
$\(A = \frac{49\sqrt{3}}{4} = \frac{7^2 \times \sqrt{3}}{4} = \frac{\left(\frac{35}{5}\right)^2 \times \sqrt{3}}{4} = \frac{\frac{35^2}{5^2} \times \sqrt{3}}{4} = \frac{35^2 \times \sqrt{3}}{25 \times 4} = \frac{35^2 \times \sqrt{3}}{100}$\)
Now, substitute this form of \(A\) into the logarithmic expression:
$\(\log_{10} A = \log_{10} \left(\frac{35^2 \times \sqrt{3}}{100}\right)$\)
Using the logarithm properties \(\log (x/y) = \log x - \log y\) and \(\log (xy) = \log x + \log y\) and \(\log x^n = n \log x\):
$\(\log_{10} A = \log_{10} (35^2 \times \sqrt{3}) - \log_{10} 100$\)
$\(\log_{10} A = \log_{10} (35^2) + \log_{10} \sqrt{3} - \log_{10} 10^2$\)
$\(\log_{10} A = 2 \log_{10} 35 + \log_{10} (3^{1/2}) - 2 \log_{10} 10$\)
Since \(\log_{10} 10 = 1\):
$\(\log_{10} A = 2 \log_{10} 35 + \frac{1}{2} \log_{10} 3 - 2$\)
Using the Given Logarithm Values
We are given \(\log_{10} 35 = 1.5441\). We need to find \(\log_{10} 3\).
We are given \(\log_{10} 1050 = 3.0212\). We can write \(1050\) as \(10 \times 105\), and \(105\) as \(3 \times 35\).
$\(\log_{10} 1050 = \log_{10} (10 \times 3 \times 35)$\)
Using the logarithm property \(\log (xyz) = \log x + \log y + \log z\):
$\(\log_{10} 1050 = \log_{10} 10 + \log_{10} 3 + \log_{10} 35$\)
Substitute the given values:
$\(3.0212 = 1 + \log_{10} 3 + 1.5441$\)
Now, solve for \(\log_{10} 3\):
$\(3.0212 = 2.5441 + \log_{10} 3$\)
$\(\log_{10} 3 = 3.0212 - 2.5441 = 0.4771$\)
This value for \(\log_{10} 3\) is a standard value often used in calculations.
Completing the Calculation
Now substitute the values of \(\log_{10} 35\) and \(\log_{10} 3\) into the expression for \(\log_{10} A\):
$\(\log_{10} A = 2 \log_{10} 35 + \frac{1}{2} \log_{10} 3 - 2$\)
$\(\log_{10} A = 2 \times (1.5441) + \frac{1}{2} \times (0.4771) - 2$\)
$\(\log_{10} A = 3.0882 + 0.23855 - 2$\)
$\(\log_{10} A = 3.32675 - 2$\)
$\(\log_{10} A = 1.32675$\)
Finally, calculate \(\log_{10} A^4\):
$\(\log_{10} A^4 = 4 \log_{10} A = 4 \times 1.32675$\)
$\(4 \times 1.32675 = 5.3070$\)
Conclusion
The value of \(\log_{10} A^4\) for the given equilateral triangle is 5.3070.
| Calculation Step | Details | Value |
|---|---|---|
| Side length \(s\) | Given | 7 cm |
| Area formula \(A\) | Equilateral triangle | \(\frac{\sqrt{3}}{4}s^2\) |
| Area \(A\) | Substitute \(s=7\) | \(\frac{49\sqrt{3}}{4}\) |
| Expression to find | Log property | \(\log_{10} A^4 = 4 \log_{10} A\) |
| Rewrite \(A\) | Using 35 and 100 | \(\frac{35^2 \times \sqrt{3}}{100}\) |
| \(\log_{10} A\) expression | Using log properties | \(2 \log_{10} 35 + \frac{1}{2} \log_{10} 3 - 2\) |
| Derive \(\log_{10} 3\) | From \(\log_{10} 1050\) and \(\log_{10} 35\) | 0.4771 |
| \(\log_{10} A\) value | Substitute values | \(2(1.5441) + 0.5(0.4771) - 2 = 1.32675\) |
| Final Value \(\log_{10} A^4\) | \(4 \times \log_{10} A\) | \(4 \times 1.32675 = 5.3070\) |
Revision Table: Key Concepts
| Concept | Description | Formula/Property |
|---|---|---|
| Equilateral Triangle Area | Area of a triangle with all sides equal (side \(s\)). | \(A = \frac{\sqrt{3}}{4} s^2\) |
| Logarithm Power Rule | The logarithm of a number raised to a power. | \(\log_b x^n = n \log_b x\) |
| Logarithm Quotient Rule | The logarithm of a division. | \(\log_b (x/y) = \log_b x - \log_b y\) |
| Logarithm Product Rule | The logarithm of a multiplication. | \(\log_b (xy) = \log_b x + \log_b y\) |
| Logarithm Base 10 | Logarithm with base 10 (\(\log_{10}\)). | \(\log_{10} 10 = 1\), \(\log_{10} 100 = 2\), etc. |
Additional Information: Logarithms and Geometry
This problem beautifully combines concepts from geometry (area of an equilateral triangle) and logarithms. Logarithms are powerful mathematical tools used to simplify complex calculations involving multiplication, division, powers, and roots.
The base-10 logarithm (\(\log_{10}\)), also known as the common logarithm, is particularly useful because our number system is base-10. This is why \(\log_{10} 10 = 1\), \(\log_{10} 100 = 2\), \(\log_{10} 1000 = 3\), and so on.
In this problem, the seemingly unrelated given values (\(\log_{10} 1050\) and \(\log_{10} 35\)) were specifically chosen to allow for the derivation of other necessary logarithm values, such as \(\log_{10} 3\). Recognizing that \(1050\) can be factored into prime numbers and related to \(35\) (\(1050 = 30 \times 35 = 10 \times 3 \times 35\)) was the key step in utilizing the given information effectively.
Similarly, rewriting the area \(A = \frac{49\sqrt{3}}{4}\) in terms of 35 (\(7 = 35/5\)) and a power of 10 (by creating 100 in the denominator) allowed us to use the given \(\log_{10} 35\) directly in the calculation of \(\log_{10} A\). This highlights how manipulating mathematical expressions is crucial for solving problems involving logarithms and specific given values.
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