Solving $x$ in exponential rate problem

Question

This is originally from a word problem in which $A$ increases at a rate of $f(x)=3^x$ and $B$ increases at a constant rate of $2x+100$. The problem asks for the point when $A$ overtakes $B$ (intercept).

How do I solve it algebraically using possibly logs or series?

Solve for $x$: $3^x=2x+100$

Thank you!

Answer 1

You need a numeric solution and should know that $3^x$ increases quickly. You can just draw the graphs to get an approximation. Maybe you know that $3^4=9^2=81$ so $x$ is a little bigger than $4$. You might try $3^{4.5}\approx 140$ and find it too large. Now you can just do bisection to get as accurate as you want.

Answer 2

Since Parcly Taxel already commented, welcome to the world of Lambert function !

The Wikipedia page will show you, for several examples, the kind of manipulations you need to arrive to the solution. Writing the equation as $$3^x=2x+a$$ would give as solution(s) $$x=-\frac{a}{2} -\frac{W\left(-\frac{ 3^{-a/2}}{2} \log (3)\right)}{\log (3)}$$ There will be two solutions corresponding to the $W_0(.)$ and the $W_{-1}(.)$ branches.

Since the argument is very small (if $a=100$ the argument is $\approx -7.65\times 10^{-25}$ ), you can use the given Taylor expansion $$W(y)=y+O\left(y^2\right)$$ and get for the $W_0(.)$ branch a value very close to $-50$.

For the $W_{-1}(.)$ branch, use the other given approximation $$W_{-1}(y)=L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(L_2-2)}{2L_1^2}+\cdots$$ where $L_1=\log(-y)$ and $L_2=\log(-L_1)$. This would lead to $$W\left(-\frac{3^{-50}}{2} \log (3)\right)\approx -59.6177$$ making $x\approx 4.26633$ while an exact solution would be $x\approx 4.26634$ (not bad, isn't it ?).

If you do not want to use Lambert function, you need, as Ross Millikan answered, some numerical method looking for the zero of $$f(x)=3^x-2x-100$$ but this function varies very fast. So, instead, consider that you look for the zero of $$g(x)=x\log(3)-\log(2x+100)$$ which looks almost as a straight line and which shows a solution just above $x=4$.

So, just use Newton method with $x_0=4$ and get the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 4.000000000 \\ 1 & 4.266349164 \\ 2 & 4.266337939 \end{array} \right)$$ which is the solution for ten significan figures.