Question 374254
I take this to mean the water left in the tank
is less than or equal to 1/10 of the starting
amount of water after {{{n}}} hours
after 1st hour: {{{y[1] - .2y[1] = .8y[1]}}}
after 2nd hour: {{{.8y[1] - .2*.8y[1] = .8^2*y[1]}}}
after 3rd hour: {{{.8^2*y[1] - .2*.8^2[1] = .8^3*y[1]}}}
The pattern here is that after {{{n}}} hours,
the amount of water left in the tank is
{{{.8^n*y[1]}}}
The problem wants to know when {{{.8^n*y[1] <= (1/10)*y[1]}}}
The {{{y[1]}}}'s cancel from both sides leaving:
{{{.8^n <= 1/10}}}
{{{10*(.8^n) <= 1}}}
Note that {{{.8^n = 8^n*10^(-n)}}}
{{{10*10^(-n)* 8^n <= 1}}}
{{{10^(-n + 1)*8^n <= 1}}}
Take the log base 10 of both sides
{{{1 - n + n*log(8) <= 0}}}
{{{n*(log(8) - 1) <= -1}}}
Now I need to multiply both sides by
{{{-1}}} which reverses the < sign
{{{n >= 1/(1 - log(8))}}}
{{{n >= 1/(1 - .90309)}}}
{{{n >= 1/.09691}}}
{{{n >= 10.3189}}}
Or, {{{n >= 10}}} in whole numbers
The answer is
{{{y[10] <= (1/10)*y[1]}}}
Unless I messed up- I think the approach is right