The problem I have is this $$y=\alpha e^{\beta x}$$
I am given the solutions y=0.5 when x = 10 and y=4 when x=50
At first I tried to approach this question like a simultaneous equation problem but since the equation is identical, after taking natural logarithms to get rid of the exponential and substituting the remaining variable cancels itself out.
Is there an approach to questions like this? It seems there is one equation and two unknowns therefore it cannot be solved by simultaneous equations.
You are right to set this up as a system of equations: $$\frac{1}{2}=\alpha e^{10\beta}$$ $$4=\alpha e^{50\beta}$$ And so we have, from the first one, $$4=8\alpha e^{10\beta}$$ and, setting the two equal, $$8\alpha e^{10\beta}=\alpha e^{50\beta}$$ $$8e^{10\beta}=e^{50\beta}$$ $$\ln 8e^{10\beta}=\ln e^{50\beta}$$ $$\ln 8+10\beta=50\beta$$ $$\beta=\frac{\ln 8}{40}$$ Great. Now, can you use this to solve for $\alpha$?