I am doing some simple models to estimate the cost of having unrealized taxed investments, but I keep running into equations that have the form of:
$$a^x = b^x\cdot(1-c) - c$$
Or even,
$$a^x = b^x\cdot d - c$$
All of the 'numbers' in the equations are in $(0,\infty)$.
From my understanding, these equations fall into the category of transcendental equations. I have no idea of to proceed to solve the above equations. I am not looking just for the solution to the equations, but also want to understand how to solve them, in case I get similar trouble in the future.
A transcendental equation contains a transcendental function that cannot be expressed in finite number of simple algebraic operations, and also such equation has no closed form expression, has no finite number of standard operations.
You can express the equation through a function $f$ and then use some simple calculus methods to at least approximate $x$
$$f(x)=b^x\cdot d - c - a^x \quad f:[0, \infty) \rightarrow \mathbb{R} \hspace{30px} \text{$a,b,c,d$ are also in the domain}$$
$$f(0)=d-c-1\hspace{30px} d\geq1+c \quad \text{or} \quad d< 1+c \\ \text{When $d=1+c$ you have $f(0)=0$}$$
$$f'(x)=d\ln{b}\cdot b^x-\ln{a}\cdot a^x \\[6px] f'(x)=0\implies a^x=d\frac{\ln{b}}{\ln{a}}b^x\\[6px]x=\log_a{d}+\log_a{\ln b}+x\cdot\log_a{b}-\log_a{\ln a}\\[6px] x=m=\frac{\log_a{d}+\log_a{\ln b}-\log_a{\ln a}}{1-\log_a{b}} \\[20px] \text{There exists only one $m$ for which the function is min or max} \\ \text{This is used for the following...}$$
You can then set $f$ to be defined by, $f(x) = d\cdot b^x -a^x$ And $g(x)=d\cdot b^x -c -a^x$
then, $f(n) = 0 \implies a^n=d\cdot b^n \implies n=\log_a{d} + n\log_a{b} \implies n=\frac{\log_a{d}}{1-\log_a{b}}$
If you think about the graph of the functions, the area between $f$ and $\frac{c-0}{t_0-n}(x-n)$ and interval $[t_0, n]$ is useful in the following
Where the second function is the line that goes through point $x:f(x)=0$, (that is point $(n, 0)$) and point $(t_0, f(n) + c)=(t_0, c)$, where $t_0: g(t_0)=0$
So $$\int_{t_0}^n f(x)-c\frac{(x-n)}{t_0-n}dx \approx 0$$
But the bigger $c$ gets, the less approximate it gets, though as $c \rightarrow 0$ both sides are equal.