Question

The pollution in a normal atmosphere is less than 0.01%. Due to leakage of a gas from a factory, the pollution is increased to 20%. If everyday 80% of the pollution is neutralized , in how many days the atmosphere will be normal( `log_2 = 0.3010`)?

A) 3
B) 4
C) 5
D) 6

Answer 1

`ln(0.0005)/ln(0.2)~=4.72` days

Explanation:

The pollution percentage is at `20%`, and we want to figure out how long it takes for it to go down to `0.01%` if the pollution decreases by `80%` every day.

This means that each day, we multiply the pollution percentage by `0.2` (`100%-80%=20%)`. If we do it for two days, it would be the percentage multiplied by `0.2`, multiplied by `0.2` again, which is the same as multiplying by `0.2^2`. We can say that if we do it for `n` days, we would multiply by `0.2^n`.

`0.2` is the original amount of pollution, and `0.0001` (`0.01%` in decimal) is the amount we want to get to. We're wondering how many times we need to multiply by `0.2` to get there. We can express this in the following equation:

`0.2*0.2^n=0.0001`

To solve it, we will first divide both sides by `0.2`:

`(cancel0.2*0.2^n)/cancel0.2=0.0001/0.2`

`0.2^n=0.0001/0.2=0.0005`

Now we can take a logarithm on both sides. Which logarithm we use doesn't really matter, we're just after the logarithm properties. I'm going to pick the natural logarithm, since it's present on most calculators.

`ln(0.2^n)=ln(0.0005)`

Since `log_x(a^b)=blog_x(a)` we can rewrite the equation:

`nln(0.2)=ln(0.0005)`

If we divide both sides, we get:

`n=ln(0.0005)/ln(0.2)~=4.72`