Solve (1 + x) to the 10th power = 1.1207.

Step-by-step solution

1. 1

   Read the equation as a growth statement. A quantity has grown by a factor of $1.1207$ (that is, $12.07\%$) over $10$ equal periods, and $x$ is the rate applied in each one. The unknown is trapped inside a $10$th power, so a root is the inverse operation.

2. 2

   Take the 10th root of both sides. $1+x = 1.1207^{1/10} = \sqrt[10]{1.1207}.$ Only the positive root is admissible, since $1+x$ must be positive for a growth factor.

3. 3

   Compute the root with natural logarithms. $\ln 1.1207 = 0.11395349$, so $\tfrac{1}{10}\ln 1.1207 = 0.01139535$ and $1+x = e^{0.01139535} = 1.01146052.$

4. 4

   Solve for x. $x = 1.01146052 - 1 = 0.01146052 \approx 1.1461\%\ \text{per period}.$

5. 5

   Verify by compounding. $1.01146052^{10} = 1.120700$, which returns the given factor, confirming the answer.

6. 6

   Compare with the 20-period rate. Halving the number of periods roughly doubles the rate: $0.00571394 \times 2 = 0.01142788$ versus the exact $0.01146052$. The tiny excess is the compounding correction - rates do not scale exactly linearly with the period count.