Analyzing the Rationality of Logarithms

The question asks us to consider two statements regarding the rationality of specific logarithmic expressions: \(\log_{10} 50\) and \(\log_{100} 10\). We need to determine if these values are rational or irrational numbers.

Statement 1: \(\log_{10} 50\) is a rational number.

Let's analyze the expression \(\log_{10} 50\). By the definition of logarithms, if \(\log_{10} 50 = x\), then \(10^x = 50\).

We can rewrite 50 as \(5 \times 10\). So, \(10^x = 5 \times 10\).

Dividing both sides by 10 (which is \(10^1\)), we get \(10^{x-1} = 5\).

This means \(x-1 = \log_{10} 5\). So, \(x = 1 + \log_{10} 5\).

For \(x\) to be a rational number, \(\log_{10} 5\) must also be a rational number (since 1 is rational, and the sum of two rationals is rational). Let's assume \(\log_{10} 5\) is rational. Then \(\log_{10} 5 = \frac{p}{q}\) for some integers \(p\) and \(q\), where \(q \neq 0\).

According to the definition of logarithm, \(10^{p/q} = 5\). Raising both sides to the power of \(q\), we get \((10^{p/q})^q = 5^q\), which simplifies to \(10^p = 5^q\).

We can write \(10^p\) as \((2 \times 5)^p = 2^p \times 5^p\). So, the equation becomes \(2^p \times 5^p = 5^q\).

By the Fundamental Theorem of Arithmetic (unique prime factorization), the prime factorization of any integer is unique. For the equation \(2^p \times 5^p = 5^q\) to hold, the powers of each prime factor on both sides must be equal.

• Comparing powers of 2: \(p\) on the left, \(0\) on the right. Thus, \(p = 0\).
• Comparing powers of 5: \(p\) on the left, \(q\) on the right. Thus, \(p = q\).

If \(p=0\) and \(p=q\), then \(q=0\). However, for \(\frac{p}{q}\) to be a rational number, the denominator \(q\) cannot be zero. This is a contradiction.

Therefore, our assumption that \(\log_{10} 5\) is rational must be false. \(\log_{10} 5\) is an irrational number.

Since \(\log_{10} 50 = 1 + \log_{10} 5\), and \(\log_{10} 5\) is irrational, \(\log_{10} 50\) is the sum of a rational number (1) and an irrational number (\(\log_{10} 5\)). The sum of a non-zero rational number and an irrational number is always irrational.

Thus, \(\log_{10} 50\) is an irrational number. Statement 1, which claims it is rational, is incorrect.

Statement 2: \(\log_{100} 10\) is an irrational number.

Let's analyze the expression \(\log_{100} 10\). Let \(\log_{100} 10 = y\). By the definition of logarithms, this means \(100^y = 10\).

We can rewrite the base 100 as \(10^2\). So, \((10^2)^y = 10\).

Using the exponent rule \((a^m)^n = a^{mn}\), we get \(10^{2y} = 10^1\).

Since the bases are the same (10), the exponents must be equal. So, \(2y = 1\).

Solving for \(y\), we find \(y = \frac{1}{2}\).

The value of \(\log_{100} 10\) is \(\frac{1}{2}\). A rational number is any number that can be expressed as the quotient or fraction \(\frac{p}{q}\) of two integers, where \(p\) is an integer and \(q\) is a non-zero integer. Here, \(p=1\) and \(q=2\), so \(\frac{1}{2}\) is clearly a rational number.

Thus, \(\log_{100} 10\) is a rational number. Statement 2, which claims it is irrational, is incorrect.

Conclusion

Based on our analysis:

• Statement 1: \(\log_{10} 50\) is a rational number - This is False.
• Statement 2: \(\log_{100} 10\) is an irrational number - This is False.

Neither of the given statements is correct.

Logarithm Rationality Analysis Summary

Revision Table: Rational and Irrational Numbers

Additional Information: Properties of Logarithms

Understanding basic logarithm properties is crucial for solving such problems:

• Definition: \(\log_b a = x\) means \(b^x = a\).
• Product Rule: \(\log_b (MN) = \log_b M + \log_b N\)
• Quotient Rule: \(\log_b (\frac{M}{N}) = \log_b M - \log_b N\)
• Power Rule: \(\log_b (M^k) = k \log_b M\)
• Change of Base Rule: \(\log_b a = \frac{\log_c a}{\log_c b}\) (for any suitable base \(c > 0, c \neq 1\))
• Special Case: \(\log_{b^k} a = \frac{1}{k} \log_b a\)

In Statement 1, we used the product rule: \(\log_{10} 50 = \log_{10} (5 \times 10) = \log_{10} 5 + \log_{10} 10\). Since \(\log_{10} 10 = 1\), this gives \(1 + \log_{10} 5\).

In Statement 2, we could use the change of base rule or the special case for base power: \(\log_{100} 10 = \log_{10^2} 10 = \frac{1}{2} \log_{10} 10 = \frac{1}{2} \times 1 = \frac{1}{2}\). Both methods confirm the result.