Find the an equation for the tangent line for the following: y=7^x at the point (1,7)

Question

I have been working on this for a while. I know I'm getting the derivative right, but I'm struggling with plugging this back into point slope form to find an equation. Here's my work, line by line.

y=ln(7)*7^x

y-7=ln(7)(7^x)(x-1)

y-7=(ln(7)x)(-ln(7))(7^x)(-7x)

y=(ln(7)x)(-ln(7))(7^x)(-7x)-7

Where am I going wrong?

Answer 1

Solution:

Your first line is wrong because if you take ln of one side you need to take ln of the other side too

such that ln y=ln 7^x = x ln 7 , x = ln y / ln7 but that will not help finding the equation of the tangent line

The function y = 7^x is of the form y =a^x and its derivative y' = a^x ln a , then y'= dy/dx = 7^xln7 is the slope of the tangent line to the function at any point. At point ( 1,7) which is the point of tangency on the curve by the tangent line, The slope of the tangent line is y' = 7^xln7 . writing this in slope form m = (y-y1) / (x-x1) ,

7^xln7= (y-7) / (x-1) ,

(x-1)7^xln7= y - 7, and y = 7^xln7(x-1) + 7 expanding, y = x7^xln7 - 7^xln 7 + 7 is the equation of the tangent line at the point (1,7)

Answer 2

Hi Chris,

I believe the mistake is in the "(-7x)" term. In expanding y-7=ln(7)(7^x)(x-1) I got the following answer:

y - 7 = ln(7) * [x*(7^x) - (7^x)]