Taking the log of an exponential function and finding the slope

[cryora]

This is part of a differential equations group project problem where I solve a set of differential equations to obtain the solution to a function. The part that I am stuck at involves taking the log of an exponential function, though there may be a mistake on the book's part, but I'm not sure.

Homework Statement

Argue from your formula for V(t), that the graph of V(t) on a log scale (i.e., the graph of log V) over an extended period of time (say, several weeks) will tend toward a graph of a straight line whose slope is either -γ (the negative reciperocol of the average lifespan of a free virus) or -μ (the negative reciprocol of the average lifespan of an infected CD4+ T cell), according to whether γ or μ is smaller.

Homework Equations

\frac{d}{dt}T(t) = λ - δT(t) - βV(t)T(t)
\frac{d}{dt}I(t) = βV(t)T(t) - μI(t)
\frac{d}{dt}V(t) = NμI(t) - γV(t)
T represents population of uninfected T cells in units of cells.
I represents population of infected T cells in units of cells.
V represents population of virus in units of virions.
λ is the rate of T cell production by the human body per day in units of 1/days
δ is the rate constant of T cells naturally dying off per day in units of 1/days
μ is the rate constant of Infected T cells dying off (bursting) per day resulting in the spread of virus in units of 1/days
γ is the rate constant of virus decaying per day in units of 1/days
N is the number of virus per cell in units of virions/cell
β is the infection rate constant in units of 1/(days*virions)

The problem states to set β = 0, assuming there is a drug that completely removes infection of T cells, allowing me to solve by 1st order linear differential equation methods to get V(t):
V(t) = \frac{NμI_0}{γ-μ}e^{-μt} + (V_0 - \frac{NμI_0}{γ-μ})e^{-γt}
with initial conditions
V(0) = V_0 \text{ and } I(0) = I_0

The Attempt at a Solution

Taking the log of V(t) and then taking the derivative, I get:
\frac{d}{dt}log[V(t)] = \frac{d}{dt}\frac{ln[V(t)]}{ln(10)} = \frac{1}{ln(10)V(t)}\frac{d}{dt}V(t)
and
\frac{d}{dt}V(t) = \frac{-Nμ^2I_0}{γ-μ}e^{-μt} - γ(V_0 - \frac{NμI_0}{γ-μ})e^{-γt}

Assuming μ is smaller than γ, the term containing e^{-γt} should go to 0 when t is large, leaving us with:
\frac{d}{dt}log[V(t)] = \frac{V'(t)}{ln(10)V(t)} = \frac{\frac{-Nμ^2I_0}{γ-μ}e^{-μt}}{ln(10)\frac{NμI_0}{γ-μ}e^{-μt}} = \frac{-μ}{ln(10)} \text{for large t}

The problem is the factor of \frac{1}{ln(10)} causing the slope of the log graph to be inconsistent with what the problem statement claims it should be: -μ.

 

[pasmith]

When doing calculus, "log" means "natural logarithm".

 

[cryora]

When doing calculus, "log" means "natural logarithm".

While that would get rid of the 1/ln10 factor, I doubt I could make that assumption in this case because for one this book does use ln notation, and there are graphs for this problem that depicts virus amount in a log base 10 scale (where it goes from 1 to 10 to 100 to... 10^n... every major gridline). The problem even states to determine μ based on the slope of those graphs.