Chapter 4 - Inverse, Exponential, and Logarithmic Functions - 4.4 Evaluation Logarithms and the Change-of-Base Theorem - 4.4 Exercises - Page 459: 77

1.13 billion years.

Given $\frac{A}{K}=0.103$, use the formula given in Example 6, we have: $t=(1.26\times10^9)\frac{ln(1+8.33(\frac{A}{K}))}{ln2}\\ =(1.26\times10^9)\frac{ln(1+8.33(0.103))}{ln2}\\ \approx1.13\times10^9\ yrs$, or 1.13 billion years.