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**Soroban** Hello, magentarita!

I got a different result . . .

At the end of $\displaystyle n$ years, Jane will have: .$\displaystyle 6\left(1.11^n\right)$ dollars.

At the end of $\displaystyle n$ years, Sarah will have: .$\displaystyle 8\left(1.08^n\right)$ dollars.

So we have: .$\displaystyle 6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}$

Take logs: .$\displaystyle \ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)$

. . Hence: .$\displaystyle n \;=\;\frac{\ln\left(\frac{4}{3}\right)}{\ln\left(\ frac{1.11}{1.08}\right)} \;=\;10.4997388$

They will have the same amount of money in about $\displaystyle {\color{blue}10\tfrac{1}{2}}$ years.

At that time, they will have: .$\displaystyle 6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\$17.95}\;\;{\color{red}(d )}$