Logarithm and an application

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My book have a really crappy proof of how

log a^x = x log a

can be true. Can someone help me?


Another question deals with an application of it which made me really confused:

Factorise (4a^3) - (29 a^2) + 47a - 10

then solve (4*4^3x) - (29*4^2x) + 47*4^x - 10 = 0


I see many similarities between the two equations, but I dunno how to factorise polynomials. Plz help me! o:)
 

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dextercioby
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Link said:
My book have a really crappy proof of how

log a^x = x log a

can be true. Can someone help me?
What other elegant proof can be given,except the one which uses the definiton??

By "log" let's see a logaritm in an arbitrary base call it {b}.Since logarithm in the base "b" and the exponential "b" to the power of "x" are inverse functions,it follows simply that:

[tex] \log_{b} a^x= \log_{b} b^{(\log_{b} a)x}=x \log_{b} a [/tex]


Link said:
Another question deals with an application of it which made me really confused:
Factorise (4a^3) - (29 a^2) + 47a - 10
then solve (4*4^3x) - (29*4^2x) + 47*4^x - 10 = 0
I see many similarities between the two equations, but I dunno how to factorise polynomials. Plz help me! o:)

It's not always possible to factorize polynomials,since that would require finding all solutions.And for polynomials of degree larger than 4,that is generally impossible.For your case,since it has integer cofficients,try to searc solutions among the integer divisors of the free term,which is "-10".
You'll "2" as a root for the polynomial.From there,just bu dividing the original polynomial through "a-2",u'll left with a second degree polynomial whose roots are easy to find.
From there u can factorize all the polynomial:To chech your result,i say that it looks like that:
[tex] P(a)=(a-2)(a-5)(a-\frac{1}{4}) [/tex]

Daniel.
 

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