Question

find an equation of the tangent line to the graph of y=log(base of 6)x at the point (1296,4)

Answer 1

y-y_1=f'(x_1)*(x-x_1)
point slope form for a line

Answer 2

@myininaya ok i used that and got 4+(1/(1296ln6))(x-1296)

Answer 3

whats the equation for y' ?

Answer 4

yeah, that looks right

Answer 5

lnx/ln6 derives to 1/xln6

Answer 6

ok becuae i didnt know if it was 4+(1/(1296ln6))(x-1296) or 4+(1/(6ln1296))(x-1296)

Answer 7

change of base rule for logs:

\[log_a(n)=\frac{ln(n)}{ln(a)}\]
youdid fine

Answer 8

thanks!

Answer 9

Alternatively,