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Thread: Year 10 Index Laws help please

  1. #1
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    Year 10 Index Laws help please

    Hey guys, I'm having a lot of struggles with expressing some problems with a prime number base, here's an examples of some questions, I have no idea how to work it, so an explanation would be very nice, Thanks!! (Btw the late X is times)
    2^x X 2
    7^t X 49

    125^x+1/5^x-1
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    Re: Year 10 Index Laws help please

    Quote Originally Posted by JamesLemming View Post
    Hey guys, I'm having a lot of struggles with expressing some problems with a prime number base, here's an examples of some questions, I have no idea how to work it, so an explanation would be very nice, Thanks!! (Btw the late X is times)
    2^x X 2
    Please do not use "x" as a variable and "X" to indicate multiplication- that is very confusing.
    Write this as "(2^x)(2). Now, do you know what "2^x" means? It is repeated multiplication of 2 isn't it? And if you multiply by another "2"? That just increases the number of "2"s by 1, doesn't it? For example, if x= 4, then 2^x= 2^4= (2)(2)(2)(2). Multiplying that by another 2 gives (2)(2)(2)(2)(2)= 2^5. The "index law", that should no be that difficult to memorize, is "(a^x)(a^y)= a^(x+y)". Here, you have (2^x)(2)= (2^x)(2^1)= 2^(x+ 1).

    7^t X 49
    You do know that 49= 7^2, don't you? Now apply what I said above to (7^t)(7^2).

    125^x+1/5^x-1
    I assume you mean 5^(x- 1). 125= 5^3. This is (5^3)(5^(x- 1)).
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    Re: Year 10 Index Laws help please

    125^x+1/5^x-1
    $\dfrac{125^{x+1}}{5^{x-1}}$ ... maybe?

    if so, use parentheses to group the terms in each exponent to make your expression clear ...
    125^(x+1)/5^(x-1)

    $\dfrac{125^{x+1}}{5^{x-1}} = \dfrac{(5^3)^{x+1}}{5^{x-1}} = \dfrac{5^{3x+3}}{5^{x-1}} = 5^{(3x+3)-(x-1)} = 5^{2x+4}$
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    Re: Year 10 Index Laws help please

    Thanks for the help guys and sorry for the misunderstanding in the question explanations, it still doesn't make any sense to me I can't see how

    (5^3)(x+1)/5^(x-1) Can go to 5(2x+4)

    If anyone could elaborate or explain, I'd be really grateful
    Thanks for all you've already done though, it's been a great help
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    Re: Year 10 Index Laws help please

    (5^3)(x+1)/5^(x-1)
    correction ... [(5^3)^(x+1)]/5^(x-1)

    exponent rule for a power to a power is $(a^x)^y = a^{xy}$, so $(5^3)^{x+1} = 5^{3(x+1)} = 5^{3x+3}$

    exponent rule for division is $\dfrac{a^x}{a^y} = a^{x-y}$, so $\dfrac{5^{3x+3}}{5^{x-1}} = 5^{(3x+3)-(x-1)} = 5^{3x+3-x+1} = 5^{2x+4}$
    Thanks from JamesLemming
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    Re: Year 10 Index Laws help please

    Oh that makes a lot of sense now!! Thank all of you for the amazing help!!
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