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    given that x>0 show that logaX^n = n logaX

    does that mean i have to learn the proof?
    i've read somewhere on my spec that says i just have to learn the law, not the explanation, but then i come across this question and the markscheme says ''see proof on page...''
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    Was this an actual exam question? If so are you sure it's a recent one from your syllabus?
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    (Original post by Darth_Narwhale)
    Was this an actual exam question? If so are you sure it's a recent one from your syllabus?
    yeah, it's in my revision book made by the exam board. im the last year to take the old spec. i must have misread the spec or made it up, but really, is that question just asking u to state the proof?
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    (Original post by ihatePE)
    yeah, it's in my revision book made by the exam board. im the last year to take the old spec. i must have misread the spec or made it up, but really, is that question just asking u to state the proof?
    Yeah it must be, I don't see what else it could be asking for...
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    (Original post by ihatePE)
    given that x>0 show that logaX^n = n logaX

    does that mean i have to learn the proof?
    i've read somewhere on my spec that says i just have to learn the law, not the explanation, but then i come across this question and the markscheme says ''see proof on page...''
    I guess you need to rewrite the proof. But it doesn't take much anyhow:

    \displaystyle \log_a(X^n) = \log_a(\underbrace{X\cdot X \cdot X \cdot X \cdot ... \cdot X}_{n \text{times}})

    =\underbrace{\log_a(X)+\log_a(X)  +\log_a(X)+...+\log_a(X)}_{n \text{times}}=n\log_a(X)
 
 
 

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