Help with deriving relationships starting with the identity a^x = e^xlna

CTil
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Hi there - it has been quite a long time since I took Calculus. I am trying to brush up and understand where to start with this question:

Starting with the identity a^x = e^xlna, derive the relationships between (a) e^x and 10^x; (b) ln x and log x. Note: log x = log10 x unless otherwise specified.

I know how the a^x was derived, but I'm honestly not sure what a) is asking? Is the 10^x regarding logarithm?
I have re-written the equation to a^x = (e^ln(a))^x
 
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Can you write ## 10 ## in the form ## e^b ##? What is ## b ##? That should help get you started.
 
here's where I am at...

f(x) = 10^x; 10^x = e^xln10; f′(x) = e^xln10 (ln 10) = 10^x ln 10
then from there I got 10^x 2.303 log 10...
 
The second part (b), in my opinion, is kind of a clumsy one in using the equation that is given. If you let ## N=a^x=e^{x \ln{a}} ##, then you can write, with ## a=10 ##, that ## x=\log_{10}(N) ## , but also ## \ln{N}=x \ln{10} ##. (Without introducing ## N ##, I think it is more difficult). Connect these last two expressions that each have an ## x ##. ## \\## Finally replace ## N ## with an ## x ##. (## N ## represents an arbitrary number in this last equation that you obtain, so you can replace it with any letter you choose. This new ## x ## is, of course, totally unrelated to the first ## x ##).
 
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So if I am following you correctly, ln N/ln 10 = log10 (N) or ln X / ln 10 = log 10 X ?
 
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CTil said:
So if I am following you correctly, ln N/ln 10 = log10 (N) or ln X / ln 10 = log 10 X ?
Correct. :smile:
 
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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