Hello Punch Originally Posted by

**Punch** Hi, can you please show me the workings to get to n> 1.523?

This is because I used logarithms to solve it. However, since this question falls under the elementary maths category, I am unable to solve it this way but I can't think of another way to solve it ...

Also, I am unable to work out the answer given by you which is [tex]

n > 1.523 by using logarithms.

Yes, you do use logarithms to solve this equation, but be careful with the direction of the inequality sign. You get:$\displaystyle (0.1)^n < 0.03$

$\displaystyle \Rightarrow n\log(0.1) < \log(0.03)$

So far, so good. But you can't then say:$\displaystyle \Rightarrow n < \frac{\log(0.03)}{\log(0.1)}$

Can you see why not? It's because you have divided both sides of the inequality by $\displaystyle \log(0.1)$ and $\displaystyle \log(0.1)$ is negative. So you must reverse the direction of the inequality, and say:

$\displaystyle \Rightarrow n > \frac{\log(0.03)}{\log(0.1)}$

If you're not sure, think about this simple example:

$\displaystyle -2x < -10$

Does that mean (dividing both sides by $\displaystyle -2$) that:

$\displaystyle x < \frac{-10}{-2}$

$\displaystyle \Rightarrow x < 5$ ?

No, of course it should be:

$\displaystyle x > 5$

Is that OK now?

Grandad