A certain substance doubles its volume every minute at 9 AM a small amount is placed in a container, at 10 AM the container was full, find the time when the container was 1/4 full

the container is 1/4 full at 9:58 AM

Step-by-step explanation:

since the volume doubles every minute, the formula for calculating the volume V at any time t is

V (t) = V₀*2^-t, where t is in minutes back from 10 AM and V₀ = container volume

thus for t=1 min (9:59 AM) the volume is V₁=V₀/2 (half of the initial one), for t=2 (9:58 AM) is V₂=V₁/2=V₀/4 ...

therefore when the container is 1/4 full the volume is V=V₀/4, thus replacing in the equation we obtain

V=V₀*2^-t

V₀/4 = V₀*2^-t

1/4 = 2^-t

appling logarithms

ln (1/4) = - t * ln 2

t = - ln (1/4) / ln 2 = ln 4 / ln 2 = 2*ln 2 / ln 2 = 2

thus t=2 min before 10 AM → 9:58 AM

therefore the container is 1/4 full at 9:58 AM