You should know that slope of the tangent is the same as the derivative at that point of contact.
so dy/dx = (ln3)(3^x)
but this equals 5 (from the line y = 5x+1)
so ln3(3^x) = 5
3^x = 5/ln3
take ln of both sides
x(ln3) = ln5 - ln(ln3)
x = (ln5 - ln(ln3)_/ln3
= ....
once you get that x, plug it into y = 3^x + 1
can somebody show me how to do this?
at what point on the graph of
y= 3^x +1 is the tangent line parallel to the line y= 5x -1 ?
1 answer
1 answer