Solve for x in each of the following equations At least 2 of the 4 must be solved algebraically. When solving graphically, sketch or paste a digital image of the graph you used to solve for the question.

- e^(2x)-e^(x)-12=0
- 2(5^(x+2))=5(2^(x+5))
- The percentage of sunlight that penetrates the water of the ocean decays exponentially with depth and can be described by the equation L(x)=100e^((k)x) where
is the % of sunlight,*L*is the depth of water in metres, and*x*is a constant that varied depending upon water conditions (*k*for clear water).*k=-0.2*- Assuming the water is clear, what percentage of the sun’s light penetrates water at 9.0m depth?
- How deep is clear water when only 5% of the sun’s light reaches it?

- A 200-g sample of a radioactive substance is placed in a chamber to be tested. After 3 h have passed only 140 g of the sample remains.
- Determine the half-life of this substance, to the nearest hundredth of an hour.
- How much of the substance will remain after 1.5 days?

Bobbly
May 15, 2018

#1**+1 **

4)

140 = 200 * (1/2)^(3/h), where h = half-life.

Divide both sides by 200

0.7 = 1/2^(3/h)

Take the natural log of both sides

3/h =Ln(0.7) / Ln(1/2)

3/h =0.514573173

Cross multiply

0.514573173h = 3

Divide both sides by 0.514573173

**h =5.83 hours - the half-life of this radioactive substance.**

1.5 days x 24 hours = 36 hours.

Use the same above formula:

R =200 * 1/2^(36/5.83)

R =200 * 1/2^(6.175)

R =200 * 0.0138401175...

**R =2.768 grams will remain after 1.5 days or 36 hours.**

Guest May 15, 2018

#2**+1 **

3) I modified your equation: e^-(xk)

L(x)=100e^-((k)x)

L(9) =100*e^-(0.2*9)

L(9) =100 * e^-1.8

L(9) =100 * 0.1653

**L(9) =16.53% - percentage of sunlight that penetrates water at a depth of 9m.**

5% = 100 * e^-(0.2x), solve for x

0.05 =100 * e^-(0.2x) divide both sides by 100

0.0005 = e^-(0.2x) take the natural log of both sides

-0.2x = -7.6 divide both sides by -0.2

**x = 38 meters - at which only 5% of sunlight will penetrate the water.**

Guest May 15, 2018