Answer:
[tex] {e}^{3v - 2u} [/tex]
Step-by-step explanation:
We have that:
[tex]u = ln(5) \: \: and \: \: v = ln(2) [/tex]
We rewrite in exponential form to get:
[tex]5 = {e}^{u} \: \: and \: \: 2 = {e}^{v} [/tex]
We rewrite 8/25 in terms of 2 and 3.
[tex] \frac{8}{25} = \frac{ {2}^{3} }{ {5}^{2} } [/tex]
This implies that:
[tex] \frac{8}{25} = \frac{ {e}^{3v} }{ {e}^{2u} } [/tex]
We use the quotient rule to get:
[tex]\frac{8}{25} = {e}^{3v - 2u} [/tex]
