L = 100*(0.9)^5d

a) substitute value d = 1.4

then L = 100*(0.9)^7

by logs

log (.9)^7 = 7log .9 = -.3203

get antilog = .478

so L = 100* .478 = 47.8

similarly for b)

Results 1 to 2 of 2

- October 16th 2008, 03:39 AM #1

- Joined
- Sep 2008
- Posts
- 9

## PLZ HELP!!!!

the amount of light as you go deeper into a lake is given by the formula L/100=(0.9)^5d, where d is the depth below the surface,in metres, and L is the percentage of the amount of light at the surface of the lake.

find the value of L when

a) d=1.4

b) d=2.7

- October 16th 2008, 04:10 AM #2

- Joined
- Oct 2008
- Posts
- 35