Understanding Exponential and Natural Log Rules: Is this Simplification Correct?

thomas49th
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Is it true that

e^{2(2x+ln2)} = e^{4x}e^{2ln2}

I can't see how that is true? According to a paper mark scheme it is.

Can someone clarify
Thanks :)
 
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First open up the brackets:
e^{2(2x+\ln 2)} = e^{4x + 2 \ln 2}
Then use a rule for e^{a + b} which you (should) know.

Actually, you can even simplify it further to 4 e^{4x}.
 
is this the same with all powers or just the exponetial function... hang on, is it a function?
 
Well, the first simplification (the one you asked about) works in general. If x is any number and a and b are two expressions, then
x^{a + b} = x^a x^b
so in particular it works for x = e, a = 4x and b = 2 ln(2).

The further simplification I spoke about just works because ln[..] is the inverse of exp[..] = e^[...]
 
yeah i can see the second one.
The first one makes sense now. Yes. I just didn't see it with the 'e'

cheerz :)
 
So the lesson to be learnt, perhaps, is that
"e^x"​
can be viewed both as the function "exp" evaluated in (some number) x, or as the number e \approx 2,7\cdots raised to the power (some number) x and that how you view it depends on the context (e.g. when differentiating it, one should view it as a function; when using simplification rules like here it's easier to just view it as an exponentiation).
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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