The number of cows is given by
[tex]14\cdot 2^{\frac{y}{5}}[/tex]
So, after [tex]k[/tex] years, the number of cows will be
[tex]14\cdot 2^{\frac{y+k}{5}}[/tex]
We want this number to be twice as much as the original:
[tex]14\cdot 2^{\frac{y+k}{5}} = 2(14\cdot 2^{\frac{y}{5}})[/tex]
First of all, we can cancel 14 from both sides:
[tex]2^{\frac{y+k}{5}} = 2\cdot 2^{\frac{y}{5}}[/tex]
Finally, on the right hand side, we can use the exponent rule
[tex]a^b\cdot a^c=a^{b+c}[/tex]
to get
[tex]2^{\frac{y+k}{5}} = 2^{\frac{y}{5}+1}[/tex]
To solve this equation, we must impose that the two exponents are the same:
[tex]\dfrac{y+k}{5} = \dfrac{y}{5}+1 \iff \dfrac{y+k}{5} = \dfrac{y+5}{5}[/tex]
And clearly this is true if and only if [tex]k=5[/tex]. So, it will take 5 years for the cow heard to double in number.
You can do the exact same steps to find the doubling time for the sheeps.
