Question

Express y as a function of \(x .\) The constant \(C\) is a positive number. \(2 \ln y=-\frac{1}{2} \ln x+\frac{1}{3} \ln \left(x^{2}+1\right)+\ln C\)

Short answer

y = \sqrt{C} \cdot (x^2 + 1)^{\frac{1}{6}} x^{-\frac{1}{4}}

Step-by-step solution

- Simplify the equation

Start with the given equation: \[ 2 \ln y = -\frac{1}{2} \ln x + \frac{1}{3} \ln (x^2 + 1) + \ln C \] Combine the logarithms on the right-hand side using the properties of logarithms.

- Use logarithm properties

Rewrite the right-hand side of the equation: \[ 2 \ln y = \ln C + \ln (x^2 + 1)^{\frac{1}{3}} - \ln x^{\frac{1}{2}} \] Simplify to: \[ 2 \ln y = \ln \left( C \cdot \frac{(x^2 + 1)^{\frac{1}{3}}}{x^{\frac{1}{2}}} \right) \]

- Exponentiate both sides

Raise both sides of the equation to the power of \(e\) to get rid of the logarithm: \[ e^{2 \ln y} = e^{\ln \left( C \cdot \frac{(x^2 + 1)^{\frac{1}{3}}}{x^{\frac{1}{2}}} \right)} \] Using the property \(e^{\ln a} = a\), we get: \[ y^2 = C \cdot \frac{(x^2 + 1)^{\frac{1}{3}}}{x^{\frac{1}{2}}} \]

- Solve for y

Take the square root of both sides to isolate \(y\): \[ y = \sqrt{C \cdot \frac{(x^2 + 1)^{\frac{1}{3}}}{x^{\frac{1}{2}}}} \] Since \(C\) is positive, \( \sqrt{C} \) is well-defined, giving: \[ y = \sqrt{C} \cdot \left( \frac{(x^2 + 1)^{\frac{1}{3}}}{x^{\frac{1}{2}}} \right)^{\frac{1}{2}} \] Finally, simplify the exponent inside the square root: \[ y = \sqrt{C} \cdot (x^2 + 1)^{\frac{1}{6}} x^{-\frac{1}{4}} \]

Key concepts

properties of logarithms

Logarithms are critical tools in mathematics, allowing one to handle exponential relationships more straightforwardly. Here are a few essential properties of logarithms that are useful for solving our problem:

• Product Property: \( \text{log}_b(M \times N) = \text{log}_b(M) + \text{log}_b(N) \)
• Quotient Property: \( \text{log}_b\frac{M}{N} = \text{log}_b(M) - \text{log}_b(N) \)
• Power Property: \( \text{log}_b(M^p) = p \times \text{log}_b(M) \)

In our problem, these properties allow us to combine and simplify multiple logarithmic terms. We used them particularly in Step 2 when we rewrote the equation:
\[ 2 \text{log}(y) = -\frac{1}{2} \text{log}(x) + \frac{1}{3} \text{log}(x^2 + 1) + \text{log}(C) \]
With the properties of logarithms, we simplified it to:
\[ 2 \text{log}(y) = \text{log}(C) + \text{log}((x^2 + 1)^{\frac{1}{3}}) - \text{log}(x^{\frac{1}{2}}) \]
And then combined the terms
\[ 2 \text{log}(y) = \text{log}\bigg( C \times \frac{(x^2 + 1)^{\frac{1}{3}}}{x^{\frac{1}{2}}} \bigg)\]Understand these properties well, as they are foundational for working with logarithmic equations and expressions.

exponentiation

Exponentiation is the process of raising a number to a power. It's the inverse operation of logarithms. In our exercise, exponentiation played a crucial role in moving from the logarithmic form to solving for the original variable.
For instance, we had:
\[ 2 \text{log}(y) = \text{log}\bigg(C \times \frac{(x^2 + 1)^{\frac{1}{3}}}{x^{\frac{1}{2}}} \bigg) \]
To eliminate the logarithm and solve for \( y \), we exponentiated both sides:
\[ e^{2 \text{log}(y)} = e^{\text{log}\bigg(C \times \frac{(x^2 + 1)^{\frac{1}{3}}}{x^{\frac{1}{2}}} \bigg)} \]
Using the property \(e^{\text{log}(a)} = a\), this simplifies to:
\[ y^2 = C \times \frac{(x^2 + 1)^{\frac{1}{3}}}{x^{\frac{1}{2}}} \]
Notice how exponentiation helped us convert a logarithmic expression into a simpler algebraic form. Subsequently, we took the square root on both sides to isolate \( y \):
\[ y = \bigg(C \times \frac{(x^2 + 1)^{\frac{1}{3}}}{x^{\frac{1}{2}}}\bigg)^{\frac{1}{2}} \]

function of a variable

A function essentially tells us how one quantity changes when another quantity changes. When we say \( y \) is a function of \( x \), denoted as \( y = f(x) \), it means the value of \( y \) depends on \( x \). In our exercise, we expressed \( y \) as a function of \( x \).
Starting with the equation:
\[ 2 \text{log}(y) = -\frac{1}{2} \text{log}(x) + \frac{1}{3} \text{log}(x^2 + 1) + \text{log}(C) \]
After simplifications and using exponential functions, we derived this form:
\[ y = \bigg(C \times \frac{(x^2 + 1)^{1/3}}{x^{1/2}}\bigg)^{1/2} \]
So now we have \( y \) explicitly defined in terms of \( x \), verifying that \( y \) indeed depends on \( x \). Breaking this down further:

• The term \( (x^2 + 1)^{1/6} \) shows how \( y \) changes with \( x^2 + 1 \). This term grows slowly because of the small exponent.
• The term \( x^{-1/4} \) shows that \( y \) changes inversely with \( x \).

Understanding how to express one variable as a function of another is crucial in many areas of math and science. It helps in analyzing how changes in one variable affect another.