Respuesta :
Answer:
[tex]y=2e^{2x}(x-\frac{\text{ln}(\frac{7}{2})}{2})+3.5[/tex]
Step-by-step explanation:
We are asked to find equation of tangent line of function [tex]y=e^{2x}[/tex] that is parallel to the linear equation [tex]y = 7x+13[/tex].
First of all, we will find the derivative of given function applying chain rule as:
[tex]y'=\frac{d}{dx}(e^{2x})\times \frac{d}{dx}(2x)[/tex]
[tex]y'=e^{2x}\times(2)[/tex]
[tex]y'=2e^{2x}[/tex]
Since derivative represents slope or rate of change, so we will equate derivative to slope of line [tex]y = 7x+13[/tex] and solve for x as:
[tex]2e^{2x}=7[/tex]
[tex]e^{2x}=\frac{7}{2}[/tex]
[tex]\text{ln}(e^{2x})=\text{ln}(\frac{7}{2})[/tex]
[tex]2x\text{ln}(e)=\text{ln}(\frac{7}{2})[/tex]
[tex]2x(1)=\text{ln}(\frac{7}{2})[/tex]
[tex]x=\frac{\text{ln}(\frac{7}{2})}{2}[/tex]
Now, we will substitute [tex]x=\frac{\text{ln}(\frac{7}{2})}{2}[/tex] in equation [tex]y=e^{2x}[/tex] to find value of y as:
[tex]y=e^{2(\frac{\text{ln}(\frac{7}{2})}{2})}[/tex]
[tex]y=e^{\text{ln}(\frac{7}{2})}[/tex]
Applying log property [tex]a^{\text{log}_a(b)}=b[/tex], we will get:
[tex]y=\frac{7}{2}=3.5[/tex]
Now, we will substitute [tex]m=2e^{2x}[/tex], [tex]x=\frac{\text{ln}(\frac{7}{2})}{2}[/tex] and [tex]y=3.5[/tex] in point slope form as:
[tex]y-3.5=2e^{2x}(x-\frac{\text{ln}(\frac{7}{2})}{2})[/tex]
[tex]y=2e^{2x}(x-\frac{\text{ln}(\frac{7}{2})}{2})+3.5[/tex]
Therefore, the equation of the tangent line would be [tex]y=2e^{2x}(x-\frac{\text{ln}(\frac{7}{2})}{2})+3.5[/tex].
