
#1
Apr412, 01:25 PM

P: 2

Hi every one, first post, so let me know if I'm not following any of the rules.
I'm studying Calculus, looking at the rules for deriving the function a^x. The first step is to change a^x to e^(x(lna)). From there, it's easy to use the chain rule to find the derivative. Why can you do that first step though? I've tried googling around, and can't find an explanation. Also, any tips on doing google searches for this kind of topic? I've tried pasting the equation into google; doing searches for "natural log" guides, "e" guides, and a browsed a few precalculus sites, but haven't found the answer I'm looking for. Thanks! 



#2
Apr412, 01:29 PM

Mentor
P: 4,499

a^{x}=(e^{ln(a)})^{x}
Do you see what to do from there? 



#3
Apr412, 01:39 PM

P: 2

Ah! I get it now.
(e^{lna}) is equal to a and (a^{b})^{c} = a^{b*c} so a^{x} = (e^{ln(a)})^{x} = e^{ln(a)*x} Thanks for the super fast reply! I feel silly for not figuring that out sooner. 



#4
Apr412, 01:45 PM

P: 418

Why does a^x = e^(x(lna))
EDIT You got it before I typed this
I think this is right, I'm just trying to remember it off the top of my head as my textbook is in school. Let the value of [itex]a^{x}[/itex] be equal to [itex]y[/itex] [itex]a^{x} = y[/itex] Take natural log of both sides [itex]ln(a^{x}) = ln(y)[/itex] Then we can bring the exponent out of the bracket [itex]x * ln(a) = ln(y)[/itex] Then we put both sides as the power of e to cancel the ln on the right [itex]e^{x * ln(a)} = e^{ln(y)}[/itex] [itex]e^{x * ln(a)} = y[/itex] Then since [itex]a^{x} = y[/itex] we sub that in for y and get [itex]e^{x * ln(a)} = a^{x}[/itex] 



#5
Apr412, 05:55 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

Another way to see the same thing is to note that [itex]aln(x)= ln(x^a)[/itex] so that [itex]e^{xln(a)}= e^{ln(a^x)}[/itex]. Then, because "[itex]f(x)= e^x[/itex]" and "[itex]g(x)= ln(x)[/itex]" are inverse functions, [itex]e^{ln(a^x)}= a^x[/itex].



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