Question

Find an equation of the tangent line to the graph of the function at the point (5, 1).
y = 8^(x − 5)

can someone please help me

Answer 1

y=8^(x-5)
ln(y)=ln(8^(x-5))
ln(y)=(x-5)ln(8)
ln(y)=xln(8)-5ln(8)
(1/y)dy/dx=ln(8)-0
dy/dx=y*ln(8)
dy/dx=8^(x-5)*ln(8)

Answer 2

hey

Answer 3

so is 8^(x-5)ln(8) my equation?

Answer 4

yes

Answer 5

oh no

Answer 6

its not the answer tho

Answer 7

o no. i thought u meant something else. thats the equation for ur derivative. u need to use it to find ur slope

Answer 8

that is the derivative right? you still need the tangent line, so you need the slope. the derivative is
\[f'(x)=8^{x-5}\ln(8)\] and
\[f'(5)=\ln(8)\]so that is your slope

Answer 9

yeah lostdove wrote "so that is my equation"

Answer 10

equation is
\[y=\ln(8)(x-5)+1\]

Answer 11

y=log(8)(x-5)+1?

Answer 12

is that correct satellite?

Answer 13

yes its correct

Answer 14

yes you want the equation of the line. you need the slope and a point, use point-slope formula
slope is ln(8) and point is (5,1)

Answer 15

point slope:
if point is \[(x_1,y_1)\]
and slope is m then
\[y-y_1=m(x-x_1)\]