To find a linear relationship between the given quantities, let's start with the equation:

$y = 3 \cdot x^{1.23} \cdot 1.2^x$

Step 1: Take the natural logarithm of both sides

To linearize this equation, apply the natural logarithm ($\ln$) to both sides:

$\ln(y) = \ln\left(3 \cdot x^{1.23} \cdot 1.2^x\right)$

Step 2: Use properties of logarithms

Expand the right-hand side using the properties of logarithms:

$\ln(y) = \ln(3) + \ln\left(x^{1.23}\right) + \ln\left(1.2^x\right)$

This further simplifies to:

$\ln(y) = \ln(3) + 1.23\ln(x) + x\ln(1.2)$

Step 3: Identify the linear relationship

Now, the equation is in a linear form:

$\ln(y) = \ln(3) + 1.23\ln(x) + x\ln(1.2)$

This equation suggests a relationship between $\ln(y)$ and both $\ln(x)$ and $x$.

• $\ln(y)$ is the dependent variable.
• $x$ and $\ln(x)$ are the independent variables.

Step 4: Rearrange in linear form

Rearrange the equation to explicitly show it as a linear relationship:

$\ln(y) = \ln(3) + 1.23\ln(x) + (\ln(1.2))x$

This can be expressed as:

$\ln(y) = (\ln(1.2))x + 1.23\ln(x) + \ln(3)$

Here:

• $\ln(1.2)$ is the slope with respect to $x$.
• $1.23$ is the coefficient of $\ln(x)$.
• $\ln(3)$ is the intercept term.

Linear Relationship

This equation represents a linear relationship between $\ln(y)$ and both $\ln(x)$ and $x$.

• If you plot $\ln(y)$ against $x$, you get a straight line with a slope of $\ln(1.2)$.
• If you plot $\ln(y)$ against $\ln(x)$, it’s linear with respect to the $\ln(x)$ term, but the overall relationship involves $x$ as well, so it’s more complex than a pure log-log plot.

Conclusion

The linear relationship between the given quantities, after applying the logarithm, is:

$\ln(y) = \ln(3) + 1.23\ln(x) + x\ln(1.2)$

Would you like more details or have any other questions?

Here are 5 related questions to deepen your understanding:

1. How can we graph this linear relationship effectively?
2. What does the slope $\ln(1.2)$ represent in this context?
3. What are some other types of transformations to linearize nonlinear equations?
4. How would this relationship change if the base of the exponent was different?
5. Can this method be used to linearize other functions involving both exponential and polynomial terms?

Tip: When dealing with logarithmic transformations, it’s important to carefully choose which variable to transform based on the type of nonlinearity you observe in the equation.