How do i solve this?

Hi anonymous,

Maybe I can help

Instead of having the original amount being y₀ units why not let it be 1 unit.  Then everything is a little easier.

So

$$\\y_n=\left( \frac{3}{4}\right)^n y_0\\\\ becomes\\ \\y_n=\left( \frac{3}{4}\right)^n \\\\$$

remember, $$y_n$$     is the amount of original water in the tank.

You want to know when this become 5%.  5% of the original 1 unit is 0.05*1=0.05

so you want to know when

  $$\left( \frac{3}{4}\right)^n=0.05\\\\$$

NOW

If you would like me to explain without changing the  $$y_0$$   to 1 ,   I can do that for you.   

The amount of old water y is found with the equation, with n being the number of refills:

  $$y_{n}=\frac{3}{4}*y_{n-1}$$

This can be changed to a version where  $$y_{n-1}$$  is irrelevant to  $$y_{n}$$ :

$$y_{n}=(\frac{3}{4})^{n}*y_{0}$$

With  $$y_{0}$$  being the starting amount of water.

We are considering percentages hence we divide  $$y_{n}$$  by  $$y_{0}$$ :

$$\frac{y_{n}}{y_{0}}=\frac{(\frac{3}{4})^{n}*y_{0}}{y_{0}}=(\frac{3}{4})^{n}$$

Now we can equate  $$\frac{y_{n}}{y_{0}}$$  with the desired percentage and the amount of required refills:

$$\frac{y_{n}}{y_{0}}=(\frac{3}{4})^{n}=0.05$$

$$n=\frac{\log{0.05}}{\log{\frac{3}{4}}}\approx10.4$$

You'll need to refill 11 times to achieve a pollution level of 5%, you can figure the second percentage out for yourself. :)

how does  $${\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)}^{{\mathtt{n}}} = {\mathtt{0.05}}$$

Thankyou Honga!