Question 1206832
Model as {{{y=ab^x}}}


{{{ab^x=y}}}
{{{5b^5=320}}}

{{{b^5=64}}}

{{{b=64^(1/5)}}}

{{{b=64^(0.2)}}}

{{{b=2.2974}}}


Model in this form is {{{highlight_green(y=5(2.2974)^x)}}}.

.

.
 {{{2.2974^x=1000/5}}}

{{{log((2.2974^x))=log((200))}}}

{{{x=log((2.2974))=log((2*100))}}}

{{{x=log((2*100))/log((2.2974))}}}

{{{x=(log((2))+log((100)))/log((2.2974))}}}

{{{x=(log((2))+2)/log((2.2974))}}}

{{{x=6.369869}}}

or x,   about 6 years 4 months