ln y= ln3 + xln2
1/y y'=ln2
y'=2*2^x ln2
so that is the slope of the line
y=slope*x+b
1.5,3*2^1.5 is a point on the line, put those in for x and y, and solve for b.
y=3⋅2^x
Find the equation of the line tangent to the graph of y at x=1.5.
2 answers
hmm... that's what i did but the internet math program says it's wrong. thanks though!
1 answer