SOLUTION: I am trying to find an equation of the tangent line at the point indicated. f (x) = log3 (8x + x-8), x = 1

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: I am trying to find an equation of the tangent line at the point indicated. f (x) = log3 (8x + x-8), x = 1      Log On


   

Question 272924: I am trying to find an equation of the tangent line at the point indicated.
f (x) = log3 (8x + x-8), x = 1
Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!
Are you sure they left that 8x+x in there for you to
add and get 9x? It seems kinda strange that they
would be testing you at this advanced level to see if
you knew how to add 8x+x and get 9x.
%22f%28x%29%22+=+log%283%2C+%288x+%2B+x-8%29%29
%22f%28x%29%22+=+log%283%2C+%289x-8%29%29
Let %22f%28x%29%22=y
y+=+log%283%2C+%289x-8%29%29
3%5Ey+=+9x-8
ln%283%5Ey%29+=+ln%289x-8%29
yln%283%29+=+ln%289x-8%29
ln%283%29%28%28dy%29%2F%28dx%29%29+=+9%2F%289x-8%29
%28dy%29%2F%28dx%29=%281%2Fln%283%29%29%289%2F%289x-8%29%29
%22f%27%28x%29%22=%28dy%29%2F%28dx%29=%281%2Fln%283%29%29%289%2F%289x-8%29%29
%22f%27%28x%29%22=%281%2Fln%283%29%29%289%2F%289x-8%29%29
%22f%27%281%29%22=%281%2Fln%283%29%29%289%2F%289%281%29-8%29%29
%22f%27%281%29%22=%281%2Fln%283%29%29%289%2F1%29
%22f%27%281%29%22=9%2F%28ln%283%29%29
m=9%2F%28ln%283%29%29
%22f%28x%29%22+=+log%283%2C+%289x-8%29%29
%22f%281%29%22+=+log%283%2C+%289%281%29-8%29%29
%22f%281%29%22+=+log%283%2C+%281%29%29
%22f%281%29%22+=+0
So the point of tangency is (1,0)
y-y%5B1%5D=m%28x-x%5B1%5D%29
y-0=%289%2F%28ln%283%29%29%29%28x-1%29
y=%289%2F%28ln%283%29%29%29%28x-1%29



Edwin