Determine the equation of the tangent to y=3(2^x) at x= 3

Question

I am not quite sure how to go about this question. If you could help that would be great.

Answer 1

Differentiating:

$$y'=3\cdot2^x\ln2$$

(Notice when differentiating a known base to a variable power, the derivative is the expression times the natural logarithm of the base)

At $x=3$, where $y'=m$ (slope):

$$y'=3\cdot2^3\ln2$$

Simplifying:

$$y'=24\ln2$$

So the slope is $24\ln2$.

Answer 2

Hint: Write $2^x \equiv e^{x \ln 2}$, then proceed to differentiate as normal to find the tangent slope. Indeed, it is a fact that $a^x \equiv e^{x \ln a}$.