# Need help in solving logarithmic equation

• Apr 2nd 2009, 04:54 PM
daunder
Need help in solving logarithmic equation
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
• Apr 2nd 2009, 05:06 PM
Soroban
Hello, daunder!

Quote:

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 $\displaystyle n$ bounces is: .$\displaystyle h \;=\;2(0.8)^n$

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

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

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

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

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

Therefore: .$\displaystyle n = 7$

• Apr 2nd 2009, 05:07 PM
daunder
Thanks for explaining (Clapping)
• Apr 2nd 2009, 05:07 PM
skeeter
Quote:

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

$\displaystyle 2(.8)^n < .5$

$\displaystyle (.8)^n < .25$

$\displaystyle n\ln(.8) < \ln(.25)$

$\displaystyle n > \frac{\ln(.25)}{\ln(.8)}$

$\displaystyle n > 6.2$

that would be 7