Financial Mathematics, Simple interest question. Help. Laurie deposits $\$60,000$ 
in a bank at $5\%$ interest per annum. 

Andrew deposits $\$40,000$ in bank at $8\%$ per annum.

How long wil it take, by simple interest, for Andrew to have more money than Laurie?

What I have done so far:

I calculated the simple interest earned by both in $5$ years:

Laurie:
$\$60,000 \cdot 5 \cdot 5/ 100 =\$15000$

Andrew:
$\$40,000 \cdot 8 \cdot 5 / 100 = \$16000$
Hint: The simple interest per year for Laruie is $0.05\cdot 60000=3000$, while the simple interest for Andrew is $0.08\cdot 40000=3200$. So the total balance of Andrew has to be greater than Laurie's, i.e. solve
$$40000+3200x&gt;60000+3000x$$
Let $L(n)$ and $A(n)$ be the amount of money Laurie and Andrew have after $n$ years, respectively. Then

$$\begin{align*}
L(n) &amp;= 60000\left(1+\frac1{20}\right)^n=60000\left(\frac{21}{20}\right)^n\\
A(n) &amp;= 40000\left(1+\frac2{25}\right)^n=40000\left(\frac{27}{25}\right)^n\\
\end{align*}$$
We want to find $$n^* = \operatorname{argmin}_n\{A(n) &gt; L(n) \}.$$

Since $L$ and $A$ are positive, increasing functions, this is equivalent to $\frac{A(n)}{L(n)}&gt;1$. We have

$$ \frac{A(n)}{L(n)} = \frac{40000}{60000}\left(\frac{27}{25}\right)^n\left(\frac{21}{20}\right)^{-n}= \frac23\left(\frac{36}{35}\right)^n$$
Since $\frac{A(n)}{L(n)}&gt;1$ if and only if $\log\left(\frac{A(n)}{L(n)}\right) &gt;0$,
$$ \log\left(\frac{A(n)}{L(n)}\right) = \log\frac23 + n\log\frac{36}{35}&gt;0\iff n&gt;\frac{\log\frac23}{\log\frac{36}{35}}\approx 14.393.$$
So Andrew will have more money after $$\lceil 14.393\rceil = 15$$ years.