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Index laws

  1. Sep 6, 2007 #1
    The index laws state that a^m x a^n = a^m+n

    These means that in the equation 6^3x-12=6^x
    6^3x/6^x = 6^2x
    but when I solve for x using logs the answer is different to what my calculator gives me. What am I doing wrong?
     
  2. jcsd
  3. Sep 7, 2007 #2

    VietDao29

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    Can you please show us what you did? Did you divide both sides by 6x?

    Well, look at it again, it's a cubic equation. :) Can you see it?

    Can you go from here? :)
     
  4. Sep 7, 2007 #3

    HallsofIvy

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    First, those are exponent laws, not "index" laws- indices are just labels distinguishing, say, x1 from x2 or x1 from x2. You can't, in general, do arithmetic with indices and there are no "laws" concerning them.

    Second, please use parentheses! I guess that when you say a^m x a^n = a^m+n, you really mean a^m x a^n= a^(m+n) because I know that is a law of exponents. But when you write 6^3x-12=6^x I don't know if you mean 6^(3x-12)= 6^x or (6^3x)-12= 6^x!

    If you mean (6^3x)- 12= 6^x, then dividing both sides by 6^x doesn't help a lot: you get (6^3x)/6^x- 12/6^x= 6^(2x)- 12(6^{-x})= 1. You could do as VietDao29 suggests: Let y= 6^x so that 6^3x= y^3 and your equation is y^3- 12= y. Similarly, you could divide by 6^x to get 6^(2x)- 12(6^{-x})= 1 and let y= 6^x in that: y^2- 12/y= 1 which is really the sames as y^3- 12= y.

    If you mean 6^(3x-12)= 6^x, then you can divide by 6^x to get 6^(2x-12)= 1 implying that 2x-12= 0 (any number to the 0 power equals 1) and x= 6.
    Conversely you could take logarithms (base 60 of both sides to get 3x-12= x whence 2x-12= 0 and, again, x= 6.
     
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