can you explain why the antiderivative of 3^x is 3^x/lnx?

Answer 1

I think the antiderivative of 3^x should be 3^x/ln(3) not ln(x), right?

As far as why....if I show that [3^x/ln(3)]' = 3^x, will that be enough?

y = 3^x/ln(3)

Take the natural log of both sides, and simplify using log rules:

ln y = ln (3^x/ln(3)) = ln(3^x) - ln(3) = xln(3) - ln(3) = ln(3)(x - 1)

Differentiate implicitly:

1/y y' = ln(3)(1)

y' = ln(3)y = ln(3)[3^x/ln(3)] = 3^x

Since we took the derivative of 3^x/ln(3) and got 3^x, the antiderivative of 3^x must be 3^x/ln(3).

Answer 2

Can you explain why the anti derivative of 3^x is 3^x/lnx?

The anti derivative (integral) of 3^x should be 3^x/ln3.

It is because
(d/dx) ( 3^x/ ln3)= (1/ln3) (d/dx) ( 3^x)
=(1/ln3) 3^x* ln 3
= 3^x

Answer 3

You can rewrite x as e^ln(x). So another way to write 3^x is as e^ln(3^x). This is also e^(x*ln(3)). If you integrate this and use the chain rule in reverse, you should wind up with the answer.

Answer 4

Given a to be a constant, a^x = exp(x ln (a)), where exp x means e^x.

Hope this helps, as this should be enough!

Answer 5

Just differentiate (3^x)/ln3, and see what you get.