Question 1148402
x, how long ago

{{{1*(1.04)^x=1200}}}

{{{1.04^x=1200}}}

{{{log((1.04^x))=log((1200))}}}

{{{x*log((1.04))=log((1200))}}}

{{{x=log((1200))/log((1.04))}}}

{{{x=181}}}{{{hours}}}
about 7.54 days before.