Just trying to find the correct, or proper, method to arrive at the author’s solution. This is the article hyperlink: Brown 1992, page 45
Here is the original equation
$$A = A_1(1 - (1 - f)e^{-a(F - T)})$$
This is the Brown’s solution, to define the term $a$
“To meet the requirement that $A$ has been approximately equal to $A_1$(equilibrium of C-14 in the upper biosphere) over the past 3500 years, we can use a trial setting of $A = 0.95A_1$ at $T = 4000$, from which”
$$a = \frac{ln(20(1 - f))}{F - 4000}$$ which approximates to
$$a ≈ \frac{2.996 + ln(1 - f)}{F - 4000}$$
This is my initial methodology
First substitute $0.95$ for the term $A$ in the original equation
$$0.95A_1 = A_1(1 - (1 - f )e^{-a(F - T)})$$
divide both sides by $A_1$, canceling out the term, resulting in
$$0.95 = 1 - (1 - f)e^{-a(F - T)}$$
add the term $((1 - f)e^{-a(F - T)})$, to both sides resulting in
$$0.95 + (1 - f)e^{-a(F - T)} = 1$$
subtract $0.95$ from both sides, resulting in
$$(1 - f)e^{-a(F - T)} = 0.05$$
since $e$ is raised to a negative exponent, the term can be expressed as $\frac{1}{e^{a(F-T)}}$ thus the equation now becomes
$$(\frac{1}{e^{a(F - T)}})(1 - f) = 0.05$$
multiply both sides by $e^{a(F - T)}$, resulting in
$$(1 - f)(1) = 0.05e^{a(F - T)}$$
From this point, two workflows can be used to derive the same solution
Workflow 1
starting with
$$(1 - f)(1) = 0.05e^{a(F - T)}$$
divide both sides by $0.05$, resulting in
$$(1 - f)(\frac{1}{0.05}) = (1)e^{a(F - T)}$$
simplify by dividing $1$ by $0.05$ and distributing $1$ to $e^{a(F - T)}$, resulting in
$$(1 - f)(20) = e^{a(F - T)}$$ take the natural logarithm of both sides
$$ln((1 - f)(20)) = ln(e^{a(F - T)})$$
use the product rule: $ln(xy) = ln(x) + ln(y)$, therefore
$$ln(1 -f ) + ln(20) = a(F - T)$$
divide both sides by $(F-T)$
$$\frac{(ln(1 -f ) + ln(20))}{(F - T)} = a$$
substitute $4000$ for $T$ and approximate $ln(20)$, this leaves the same solution as Brown 1992
Workflow 1 Solution $$\frac{(ln(1 -f ) + 2.996)}{(F - 4000)} = a$$
Workflow 2
beginning with
$$(1 - f)(1) = 0.05e^{a(F - T)}$$
distribute $(1)$ to $(1 - f)$ resulting in
$$(1 - f) = 0.05e^{a(F - T)}$$
divide both sides by $0.05$
$$\frac{(1 - f)}{0.05} = e^{a(F - T)}$$
take the natural log of both sides
$$ln\frac{(1 - f)}{0.05}) = ln(e^{a(F - T)})$$
use the quotient rule: $ln(x/y) = ln(x) - ln(y)$, thus
$$ln(1 - f) - ln(0.05) = a(F - T)$$
divide both sides by $(F - T)$
$$\frac{(ln(1 -f ) - ln(0.05))}{(F - T)} = a$$
approximate $ln(0.05)$
$$\frac{(ln(1 -f ) - (-2.996))}{(F - T)} = a$$
simplify, note the sign change, then substitute $T = 4000$
Workflow 2 Solution $$\frac{(ln(1 -f ) + 2.996)}{(F - 4000)} = a$$
Both Workflow 1 and 2 equate to $$\frac{(ln(1 -f ) + 2.996)}{(F - 4000)} = a$$
Which is the same as Brown’s solution for $a$ after he approximated $ln(20)$ and applied the product rule:
$$a ≈ \frac{2.996 + ln(1 - f)}{F - 4000}$$
I’m just not sure which of my workflows is correct, they both may be wrong, as algebra is not my strong suit. I think that there may be a problem with distributing (1) to (1-f) in workflow 2 might be wrong, it might be just chance that the sign change from using the quotient rule is just a lucky coincidence since $ln(0.05) ≈ -2.996$ and $ln(20) ≈ 2.996$.
Anyhow, I’m sure that I did something(s) wrong here. Any advice on my solutions and workflows that the community can give would be of great help.
Thank you in advance,
Mathmoot
PS: I'm sure that I went overboard on the parenthesis.
Both are fine.
Anything multiplied by $1$ is itself as $1$ is the multiplicative identity.
Also $\frac1{20}=0.05$ and hence $-\ln 20 = \ln 0.05$.