y=3⋅2^x

Find the equation of the line tangent to the graph of y at x=1.5.

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.

hmm... that's what i did but the internet math program says it's wrong. thanks though!

To find the equation of the line tangent to the graph of y at x=1.5, we need to find the slope of the tangent line and the point of tangency.

First, we differentiate the given equation to find the derivative, which will give us the slope of the tangent line at any point on the graph.

The derivative of y = 3⋅2^x can be found using the power rule. The power rule states that if the equation is in the form y = a⋅b^x, then the derivative of y with respect to x is given by dy/dx = a⋅b^x⋅ln(b), where ln(b) is the natural logarithm of b.

In this case, a = 3 and b = 2. So, the derivative of y = 3⋅2^x is:

dy/dx = 3⋅2^x⋅ln(2)

Now we have the derivative, which gives us the slope of the tangent line at any point on the graph. To find the slope at x=1.5, we substitute x=1.5 into the derivative:

dy/dx = 3⋅2^1.5⋅ln(2)

Now we can find the slope of the tangent line at x=1.5.

Next, we need to find the point of tangency. To do this, substitute x=1.5 into the original equation:

y = 3⋅2^1.5

Now we have the x-coordinate (x=1.5) and the y-coordinate of the point of tangency.

Finally, we can use the slope-intercept form of a line, y = mx + b, and substitute the values we found to obtain the equation of the tangent line. The slope m is the value we calculated earlier, and (x, y) are the coordinates of the point of tangency that we found.

Therefore, the equation of the line tangent to the graph of y at x=1.5 is:

y = (3⋅2^1.5⋅ln(2))(x - 1.5) + 3⋅2^1.5