C3 Maths help

As an exercise, see if you can prove the following result:
$\frac{d}{dx}a^x = a^x\ln{x}$ (a is a real number)
It follows that if you find the x such that lnx=1, you get backwhat

Hmm, I'm assuming the OP is doing Edexcel(If I'm wrong I apologise) . Exponential and natural logs comes before the differentiation chapter.

$\frac{d}{dx} a^x = a^x lna$ is a C4 concept.

The power series for e^x is a FP3 concept (probably soon to change).

So talking about these are probably not in the best interest of the OP if they're struggling to comprehend what the function e^x and lnx is.

Hrov: Firstly I would ask you to read your textbook, and then come back and ask questions on things you don't understand from that.

Im doing aqa c3

(I deleted my last message)

e is just a number. So the graph of e^x will be just like y=2^x or y=7^x. e has some special properties though, and through some work you can find that when you differentiate e^x you get e^x back (unlike with, say, 3^x or 194^x).

It's easy to forget that the exponential function is just the graph y = 2.7818...^x

The log function lnx is the just inverse of the exponential function e^x. So you'll have:
e^(lnx) = x, and
ln(e^x) = xlne = x

e^x = 7
ln both sides:
ln(e^x) = ln7
Use your logarithm rules:
xlne = ln7
Use the fact that lne=1 **:
x = ln7

** Because e^x and lnx are inverses, you can draw the graph of lnx by just drawing the mirror of y=e^x in the line y=x. From the graph y=lnx it then becomes "obvious" why lne=1 - because the point (1,e) is on the graph y=e^x, the point (e,1) will be on the graph y=lnx.