Calculating Log Y Using Logarithm Properties

The problem provides us with two pieces of information: the value of log x and an equation relating y to x. We are given that \( \text{log x} = 1.25 \) and \( y = x^{\text{log x}} \). Our goal is to find the value of \( \text{log y} \).

To solve this, we need to use the properties of logarithms, specifically the power rule.

Step-by-Step Process to Find Log Y

Let's follow these steps using the given information:

1. We are given the equation \( y = x^{\text{log x}} \).
2. To find \( \text{log y} \), we take the logarithm of both sides of this equation. Assuming the base of the logarithm is consistent (it doesn't matter what the base is, as long as it's the same for both \( \text{log x} \) and \( \text{log y} \)), we get: \( \text{log y} = \text{log}(x^{\text{log x}}) \)
3. Now, we apply the logarithm power rule, which states that \( \text{log}(a^b) = b \cdot \text{log a} \). In our expression \( \text{log}(x^{\text{log x}}) \), the base is x and the exponent is log x. So, we can bring the exponent down as a multiplier: \( \text{log y} = (\text{log x}) \cdot (\text{log x}) \)
4. We are given that \( \text{log x} = 1.25 \). Substitute this value into the equation: \( \text{log y} = (1.25) \cdot (1.25) \)
5. Finally, calculate the product of 1.25 and 1.25.

Calculating the Product of 1.25 and 1.25

We need to compute \( 1.25 \times 1.25 \).

This is the same as squaring 1.25.

\( 1.25 \times 1.25 = 1.5625 \)

To perform the multiplication, you can think of it as:

Multiply 125 by 125:

\( 125 \times 125 = 15625 \)

Since there are two decimal places in 1.25 and two decimal places in the other 1.25, the result will have \( 2 + 2 = 4 \) decimal places.

So, \( 1.25 \times 1.25 = 1.5625 \).

Alternatively, using the concept of fractions or algebraic expansion:

\( 1.25 = 1 + 0.25 = 1 + \frac{1}{4} = \frac{5}{4} \)

\( 1.25 \times 1.25 = \frac{5}{4} \times \frac{5}{4} = \frac{5 \times 5}{4 \times 4} = \frac{25}{16} \)

To convert \( \frac{25}{16} \) to a decimal, divide 25 by 16:

\( 25 \div 16 = 1 \text{ with a remainder of } 9 \)

\( \frac{9}{16} = 0.5625 \)

So, \( 1 + 0.5625 = 1.5625 \).

Result for Log Y

From our calculation, we found that:

\( \text{log y} = 1.5625 \)

This is the value of log y.

Revision Table: Key Logarithm Properties

Additional Information on Logarithm Calculations

When working with logarithms, understanding their relationship with exponents is fundamental. A logarithm answers the question "What exponent do I need to raise the base to, to get this number?". For example, \( \text{log}_{10}(100) = 2 \) because \( 10^2 = 100 \).

In this problem, the base of the logarithm is not explicitly stated for \( \text{log x} \) or \( \text{log y} \). However, the power rule works for any valid logarithm base \( b > 0, b \neq 1 \). Since we were given the value of \( \text{log x} \) directly and used it in a multiplication, the specific base does not affect the final numerical answer for \( \text{log y} \), as long as the base used for \( \text{log x} \) is the same as the base for taking the log of y.

Problems like this demonstrate how applying logarithm properties can transform expressions and make calculations possible when direct evaluation might be difficult.