How to bring $\log(y) = \log(a) + \log(e^{bx})$ to $y=mx+c$ format?

Question

I was solving a problem regarding equation of the straight line. The question is to find the best value of $a$ and $b$ if

$$ y = ae^{bx}. $$

What I tried?

I multiplied LHS and RHS with $\log$ in order to separate $a$ and $b$ and as a result I got the below equation.

$$ \log(y) = \log(a) + \log(e^{bx}).$$

But I am stuck.

how should I proceed to convert this equation to $y=mx+c$ format. Any help and guidance is much appreciated. Thanks.

Answer 1

\begin{align} y &= ae^{bx}\\ \implies \ln(y) &=\ln(ae^{bx})\\ \implies \ln(y) &=\ln(a) + \ln(e^{bx})\\ \implies \ln(y) &= \ln(a) + bx\\ \implies \color{green}{\ln(y)} &= \color{blue}{b}x + \color{red}{\ln(a)}\\ \implies \color{green}{Y} &= \color{blue}mx + \color{red}{C} \end{align}

Answer 2

you are so close.

$$ y = a\mathrm{e}^{bx} $$ taking logs have you have done $$ \log(y) = \log\left(a\mathrm{e}^{bx}\right) = \log(a) + \log\left(\mathrm{e}^{bx}\right) = \log(a) + bx $$ you did not apply the log fully on the exponential

Now you can relabel $$ \log(y) \to y'\\ \log(a) \to c $$ we now have $$ y' = bx + c $$ which we now have the form you desire.