The thing is to solve for n.
This is how far I got: $FV=PV(1-d)^{-t}$
So, $1000(1-0.05)^{-10} + 2000(1-0.05)^{-5}= 1000(1-0.1)^{-(10-n)}+2000(1-0.1)^{-(10-2n)}$
$\Longrightarrow\;\; 2584.71=1000(0.9)^{-(10-n)}+2000(0.9)^{-(10-2n)}$
$\Longrightarrow\;\; 2.58471= (0.9)^{-10}(0.9)^n + 2(0.9)^{-10}(0.9)^{2n}$
$\Longrightarrow\;\; 2.58471=2.87(0.9)^n+5.74(0.9)^{2n}$
let $X=0.9^n$ we get: $5.74X^2+2.87X-2.58471$ Solve for $X=0.47$
$\Longrightarrow\;\; 0.47=0.9^n$ then applied $\ln$ to both sides and got $n=7$.
Is this correct?!
Fund 1: discount $d=5\%$, Fund 2: interest $i=10\%$.
$$ \frac{1000}{(1-d)^{10}}+\frac{2000}{(1-d)^{5}}=1000(1+i)^{10-n}+2000(1+i)^{10-2n}\tag 1 $$
$$ 4,254.89=\underbrace{1000(1.1)^{10}}_{2,593.74}\left(1.1^{-n}+2\times 1.1^{-2n}\right)\tag 2 $$ Putting $x=1.1^{-n}$ and $a=\frac{4,254.89}{2,593.74}=1.64$ the $(2)$ becomes $$ x+2x^2=a \tag 3 $$ and solving equation $(3)$ we find the roots $$ x_{1,2} = \frac{\pm\sqrt{8 a+1}-1}{4} \quad\Longrightarrow x_1=-1.18941,\;x_2= 0.689415 $$ Discarding the negative solution we have $1.1^{-n}=0.689415$ and then $-n\log (1.1)=\log{0.689415}$, that is