Solving for $$ y $$ after the transformation $$ (x, y) \rightarrow(\log x, \log y) $$ produces exponential model power model logarithmic model

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When you apply the transformation $$(x, y) \rightarrow (\log x, \log y)$$, you're essentially linearizing a relationship that may originally be nonlinear. Let's explore what kind of model this transformation leads to.

### Understanding the Transformation

1. **Original Model**: Suppose the original relationship between $$ x $$ and $$ y $$ is a **power law**:
   $$
   y = a \cdot x^b
   $$
   
2. **Applying Log Transformation**:
   $$
   \log y = \log (a \cdot x^b) = \log a + b \cdot \log x
   $$
   
   This equation is linear in terms of $$\log x$$ and $$\log y$$:
   $$
   \log y = b \cdot \log x + \log a
   $$
   
   Here, $$ \log y $$ is a linear function of $$ \log x $$, with slope $$ b $$ and intercept $$ \log a $$.

### Resulting Model

After applying the log transformation and solving for $$ y $$, you revert back to the original power law form:
$$
y = a \cdot x^b
$$

This indicates that the transformation has **linearized a power model**. Therefore, the resulting model after the transformation is best described as a **power model**.

### Summary

- **Transformation Applied**: $$(x, y) \rightarrow (\log x, \log y)$$
- **Resulting Model**: Power Model
- **Equation**: $$ y = a \cdot x^b $$

### Final Answer

**A power law model is obtained by solving for $$ y $$ after this logarithmic transformation.**
After transforming $$(x, y) \rightarrow (\log x, \log y)$$, the resulting model is a **power model** of the form $$ y = a \cdot x^b $$.

Solving for \( y \) after the transformation \( (x, y) \rightarrow(\log x, \log y) \) produces exponential model power model logarithmic model

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