Question 239293
Some keys to understanding how to do this problem:<ul><li>Using percents in formulas or functions is not very practical. So, since 75% is 3/4, we will use 3/4 instead of 75%. (We could also use 0.75 instead of 75%.)</li><li>When problems say "75% of" some number (or "3/4 of") some number, that "of" translates into a multiplication. The beer-lambert-bougar law says that the intensity of light is some percent of the intensity of the light 1 meter above. So to find the intensity of light in your lake, we will multiply the intensity of light above by 75% (or 3/4).</li></ul>
So to figure out the answers to your problem, let's make up a number for the intensity of light just as it hits the water (i.e. the depth is zero):<ul><li> Let's say the intensity at depth 0 is 100.</li><li>The intensity at a depth of 1 meter would be 3/4 of intensity of the light at 0 meters: {{{100(3/4)}}}.</li><li>At 2 meters depth the intensity would be 3/4 of the intensity at 1 meters depth: {{{100(3/4)(3/4)}}}</li><li>At 3 meters depth the intensity would be 3/4 of the intensity at 2 meters depth: {{{100(3/4)(3/4)(3/4)}}}</li><li>etc.</li></ul>
a) Explain why intensity I is an exponential function of depth d in meters
We can see that the intensity is calculated, in part, by the repeated multiplication of 3/4. Our formula is an exponential function because we will use an exponent to represent the repeated multiplication of 3/4.<br>
b) Use a formula to express intensity I as an exponential function of d. (use I0 to denote the initial intensity.)
From the example above we can see where the initial intensity (intensity at depth 0) fits in the formula:
{{{I = I[0](3/4)^d}}}
where I is the intensity at d meters, {{{I[0]}}} is the initial intensity and d is the depth in meters.<br>
c) Explain in practical terms the meaning of {{{I[0]}}}.
It is the intensity at depth 0 meters<br>
d) At what depth will the intensity of light be one tenth of the intensity of light striking the surface?
One tenth of the intensity of light striking the surface is {{{(1/10)I[0]}}} (Remember that "a fraction of a number" means that fraction times that number.) So the equation we need to solve is:
{{{(1/10)I[0] = I[0](3/4)^d}}} 
First we'll divide by {{{I[0]}}}
{{{((1/10)I[0])/I[0] = (I[0](3/4)^d)/I[0]}}} 
On both sides the {{{I[0]}}}'s cancel leaving:
{{{1/10 = (3/4)^d}}}
To solve for d, when it is the exponent like this, we will use logarithms:
{{{log((1/10)) = log(((3/4)^d))}}}
The left side is -1 because 1/10 is {{{10^-1}}}. (If this is not obvious, use your calculator. So now we have:
{{{-1 = log(((3/4)^d))}}}
On the right side we can use the property of logarithms, {{{log(a, (p^q)) = q*log(a, (p))}}}, to move the exponent from the argument out in front of the logarithm giving us:
{{{-1 = d*log((3/4))}}}
And now we can divide both sides by {{{log((3/4))}}}:
{{{((-1)/log((3/4))) = ((d*log((3/4)))/log((3/4)))}}}
On the right side the {{{log((3/4))}}}'s cancel leaving:
{{{((-1)/log((3/4))) = d}}}
Using our calculator on the remaining log:
{{{((-1)/-0.1249387366083000) = d}}}
And using the calculator to divide we get:
{{{8.0039227796510935 = d}}}
So at very close to 8 meters the intensity of light will be 1/10 of what it was at the surface.