NUMBAT OER - Open Educational Resources

6. Negative exponents

If you plot an exponential growth curve, it is clear that the slope (gradient) of the curve increases as the independent variable (for instance time) increases. The rate at which the slope changes is determined by the constant, denoted k in previous sections.

If k is negative, the dependent variable decreases as the independent variable increases. Instead of increasing, the slope of a negative exponential curve decreases as the independent variable increases, so that the curve appears gradually to level out. An example is shown here:

Example negative exponential function

An example of exponential decrease (or decay) is the absorption of light by a medium such as water. Imagine that each photon has a defined chance of being absorbed as it traverses a metre of water. If this chance is 5% (1 in 20), 95% of the photons will pass through a metre of water. Over the next metre, 5% of the remaining photons will be absorbed, leaving 90.3% of the photons to enter the third metre layer. As the number of photons decreases, the number absorbed over each successive metre also decreases.

We can write a relationship between light level and depth as an exponential function:

Ez = E0.exp(-KD.z)

where E0 and Ez are the light intensity at the surface and a depth of z metres, respectively, and KD is the absorption coefficient (more properly the 'diffuse attenuation coefficient', in units of depth-1).

An important quantity in the functioning of aquatic ecosystems is the depth to which sufficient light penetrates to allow photosynthetic growth to take place. Although many different factors influence the minimum amount of light plants need for photosynthesis, a good rule of thumb at least for the open ocean is that photosynthetic growth is possible down to 1% of the light incident at the surface. If we term this depth z1% and call the 1% light level E1%, we can rewrite the equation:

E1% = E0.exp(-KD.z1%)

We can rewrite E1% in terms of E0, and then put this value into the equation and re-arrange this to give an equation relating depth and KD:

E1% = 0.01 . E0

0.01 . E0 = E0.exp(-KD.z1%)

exp(-KD.z1%) = 0.01 . E0/E0 = 0.01

-KD.z1% = ln(0.01)

The symbol 'ln' indicates the natural logarithm (that is the logarithm expressed to the base e). So it is now possible to calculate the value of KD for a known 1% 'light depth', or the 1% depth if you know KD:

KD = -ln(0.01)/z1%

z1%= -ln(0.01)/KD

[Technical note: the logarithm of a number less than one has a negative sign. KD is conventionally written as a positive number, so the minus sign in the exponent provides an answer in 'positive' metres.]

For a 1% light depth of 50 m, the value of KD is 0.092103 m-1, whereas for a 1% light depth of 20 m KD is 0.230258 m-1 (note the units of 'per metre'). The diffuse attenuation coefficient increases as water becomes more turbid (ie absorbs more light), corresponding to a shallower 1% light depth. In the initial discussion of light penetration, we suggested that photons had a 5% chance of being absorbed within one metre. Thus KD is 0.05, and the 1% light level occurs at a depth of 92 m – the sort of value that is only found in very clear water such as the middle of deep ocean areas.

You can use the same basic equation to work out the proportion of surface light that reaches a given depth, z* if you know the value of KD:

Ez* = E0.exp(-KD.z*)

Ez*/ E0 = exp(-KD.z*)

Where the expression Ez*/E0 is the proportion of surface light at a depth of z*. If you wanted to express this as a percentage, you would need to multiply the result by 100.