To prove that \( z = \frac{xy}{x+y} \) given \( 3^x = 4^y = 12^z \), follow these steps:

### Step 1: Express the equality in terms of a common variable
Let \( k = 3^x = 4^y = 12^z \). Then we have:
\[
k = 3^x = 4^y = 12^z
\]
Taking the logarithm (in any base, but typically natural logarithms) of all sides gives:
\[
\ln(k) = x \ln(3) = y \ln(4) = z \ln(12)
\]

### Step 2: Solve for \( x \), \( y \), and \( z \) in terms of \( \ln(k) \)
From the logarithmic expressions:
\[
x = \frac{\ln(k)}{\ln(3)}, \quad y = \frac{\ln(k)}{\ln(4)}, \quad z = \frac{\ln(k)}{\ln(12)}
\]

### Step 3: Use the relationship between \( \ln(12) \), \( \ln(3) \), and \( \ln(4) \)
Since \( 12 = 3 \times 4 \), the logarithm of 12 can be written as:
\[
\ln(12) = \ln(3 \times 4) = \ln(3) + \ln(4)
\]
Thus, we can rewrite \( z \) as:
\[
z = \frac{\ln(k)}{\ln(3) + \ln(4)}
\]

### Step 4: Express \( z \) in terms of \( x \) and \( y \)
Substitute the expressions for \( x \) and \( y \) from Step 2:
\[
\frac{1}{z} = \frac{\ln(3) + \ln(4)}{\ln(k)} = \frac{\ln(3)}{\ln(k)} + \frac{\ln(4)}{\ln(k)} = \frac{1}{x} + \frac{1}{y}
\]

Taking the reciprocal of both sides:
\[
z = \frac{xy}{x + y}
\]

### Conclusion
We have proven that \( z = \frac{xy}{x + y} \) given that \( 3^x = 4^y = 12^z \).

Would you like further details or have any questions?

Here are some related questions:
1. How does this proof change if the base numbers are different?
2. What if we had an equation of the form \( a^x = b^y = c^z \)?
3. Can this be extended to more complex exponential relationships?
4. What is the significance of the relationship \( \ln(12) = \ln(3) + \ln(4) \) in this proof?
5. How can logarithmic identities help in solving other exponential equations?

**Tip:** When dealing with exponential equations, logarithms can simplify the process of comparing exponents by turning multiplicative relationships into additive ones.

Explore how to prove z = xy / (x + y) given the equations 3^x = 4^y = 12^z using logarithmic identities and exponential equations. This problem involves advanced concepts in exponential functions and logarithms, suitable for advanced high school and college students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation