maths differentiation question help

A curve has the equation y=ln3x - e^-2x
Show that the equation of the tangent at the point with an x-coordinate of 1 is y=((e^2+2)/e^2)x - ((e^2+3)/e^2)+ln3

Please help I have no idea how to do this question. Help is much appreciated thank you

It's always the same process to find the equation of a tangent to a curve at a point, you need the coordinates of the point and the gradient of the tangent to the curve at that point, which you find by differentiating. (I hope this is obvious, but you do say you have "no idea" how to do the question, so I thought I'd check.)

Assuming you're OK with that, it should be easy to find the y-coordinate to go with the x-coordinate, so I'm assuming it's the differentiation that's the issue. Is it one of the pieces you have to differentiate that is being tricky? Do you know how to differentiate lx(3x)? How about e^-2x?

Yes i know to differentiate that but just seems to not getting the final right answer.

Differentiation: dy/dx= 1/x + 2e^-2x

then, i sub in x=1 into the dy/dx but i got 1.270670566

after that, i sub in the 1.270670566 into the equation of y=ln3x - e^-2x and got 1.259...

When i put them values into the format of y-y1=m(x-x1), i did not get the final answer as shown in the question so Idk where it went wrong.

Why are you rounding your values when the answer you seek has nothing rounded at all??

Just work with exact values. Also, why are you subbing in the dy/dx value back into the y equation ??

how do you find y-value?

Sub in x=1 into the equation for y.

Yes i know to differentiate that but just seems to not getting the final right answer.

Differentiation: dy/dx= 1/x + 2e^-2x

then, i sub in x=1 into the dy/dx but i got 1.270670566

after that, i sub in the 1.270670566 into the equation of y=ln3x - e^-2x and got 1.259...

When i put them values into the format of y-y1=m(x-x1), i did not get the final answer as shown in the question so Idk where it went wrong.