Find an equation for the tangent line of $y = e^{4x}$ that is parallel to the linear equation $y = 5 x + 17$. I don't know how to solve this problem:

Find an equation for a tangent line of $y = e^{4x}$ that is parallel to the line given by $y = 5x + 17$.

I have tried with multiple approaches and have gotten 
$$y-e^4(4e^{20})=5(x- 4e^{20}).$$
And my answer remains incorrect any help would be greatly appreciated
 A: The tangent line will have gradient $5$, so you have to find an $x$ such that 
$\frac{d (e^{4x})}{dx} = 5$. Since $\frac{d (e^{4x})}{dx} = 4e^{4x}$, setting
$x = \frac{\log\left(\frac{5}{4}\right)}{4}$ works. Since $\frac{dy}{dx} = 4y$ when $y = e^{4x}$, then $y$ will be $\frac{5}{4}$ when the gradient is $5$.
The line is $y = 5x + c$, where $y$ is $\frac{5}{4}$ when $x$ is $\frac{\log\left(\frac{5}{4}\right)}{4}$. Plugging these values in
gives $\frac{5}{4} = 5\left(\frac{\log\left(\frac{5}{4}\right)}{4}\right) + c$. This means that $c = \frac{5}{4}\left(1 - \log\left(\frac{5}{4}\right)\right)$. So the line is $y = 5x + \frac{5}{4}\left(1 - \log\left(\frac{5}{4}\right)\right)$
A: Essentially you are trying to find the equation of the tangent line of $y=e^{4x}$ that is parallel to the line $y=5x+17$. This equates to finding when the derivative of your function is equal to 5. With a bit of calculus you can find that this happens whenever $x=\frac{1}{4}\ln(\frac{5}{4})$ which means that $y=\frac{5}{4}$. Now your line must have slope $m=5$ so that it will be parallel to $y=5x+17$. Therefore, $y=5x+b$ and to solve for b we use the point $(\frac{1}{4}\ln(\frac{5}{4}),\frac{5}{4})$. Finally your line should be: $$ y = 5x+\frac{5}{4}(1-\ln(\frac{5}{4}))$$
A: Think about what the problem means.
Two lines are parallel if they have the same slope.
The slope of the tangent line is equal to the derivative at that point.
So you are to find the equation of a line where $\frac {dy}{dx}$ is equal to the slope of $y = 5x + 17$.
The slope of $y = 5x + 17$ is $5$.  (It's in the slope intercept form $y = mx + b$ so the slope is $m$).
So we need to find the value $x_1$ frac $\frac {dy}{dx} = 5$.
And so we need to equation of a line that has slope $=5$ and contains the point $(x_1, y_1)$.
Can you do that?
.......
By the chain rule:  $\frac {de^4x}{dx} = \frac {de^{4x}}{d4x}\frac {d4x}{dx}= e^{4x}\dot 4=4e^{4x}$.
So we need to solve $4e^{4x_1} = 5$.
$e^{4x_1} =\frac 54$
$4x_1 = \ln \frac 54$
$x_1 = \frac {\ln\frac 54}{4}$.  And the point on the graph $y=e^{4x}$ where this occurs is 
$y_1 = e^{4x_1} =e^{4\frac {\ln\frac 54}{4}}=e^{\ln \frac 54} = \frac 54$.
The tangent line of $y=e^{4x}$ at $(\ln \frac 54, \frac 54)$ has a slope of $5$.
We need to find the equation of a line containing the point $(x_1, y_1)=(\ln \frac 54, \frac 54)$ and has a slope of $5$.
.....
So 
$m = 5 = \frac {y-y_1}{x-x_1}$ so $y-y_1 = 5(x-x_1)$ so $y = 5(x-x_1) + y_1= 5x +  [y_1 - 5x_1]$.
SO the equation for the line is:
$y = 5x + (\frac 54 - 5\cdot\ln \frac 54)$.
