Need help in solving logarithmic equation

**daunder** · April 2nd 2009, 04:54 PM

A ball is made of rubber, when dropped on a concrete floor, bounces up to 80% of the height from which it was dropped. The ball is dropped from 2m. After how many bounces will the ball not rise over 0.5 m?

This is what I did

.5 = 2(.8) ^n

.25 = .8^n

ln .25 = n ln .8

n = 6.2

Now, what would be the answer 6 drops or 7 drops

**Soroban** · April 2nd 2009, 05:06 PM

Hello, daunder!

A ball is made of rubber, when dropped on a concrete floor,
bounces up to 80% of the height from which it was dropped.
The ball is dropped from 2m.
After how many bounces will the ball not rise over 0.5 m?

The height after $n$ bounces is: . $h \;=\;2(0.8)^n$

We want an inequality: . $2(0.8)^n \:\leq \:0.5$

Then: . $(0.8)^n \:\leq \:0.25$

Take logs: . $\ln(0.8^n) \:\leq \:\ln(0.25) \quad\Rightarrow\quad n\!\cdot\!\ln(0.8) \:\leq \:\ln(0.25)$

Note: . $\ln(0.8)$ is a negative quantity.

Divide by $\ln(0.8)\!:\;\;n \:\geq \:\frac{\ln(0.25)}{\ln(0.8)} \:=\: 6.2125...$

Therefore: . $n = 7$

**daunder** · April 2nd 2009, 05:07 PM

Thanks for explaining

**skeeter** · April 2nd 2009, 05:07 PM

Originally Posted by daunder

A ball is made of rubber, when dropped on a concrete floor, bounces up to 80% of the height from which it was dropped. The ball is dropped from 2m. After how many bounces will the ball not rise over 0.5 m?

This is what I did

.5 = 2(.8) ^n

.25 = .8^n

ln .25 = n ln .8

n = 6.2

Now, what would be the answer 6 drops or 7 drops