1. ## 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

2. ## Re: Year 10 Index Laws help please

Originally Posted by JamesLemming
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)).

3. ## 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)

$\dfrac{125^{x+1}}{5^{x-1}} = \dfrac{(5^3)^{x+1}}{5^{x-1}} = \dfrac{5^{3x+3}}{5^{x-1}} = 5^{(3x+3)-(x-1)} = 5^{2x+4}$

4. ## 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

5. ## 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 $\dfrac{5^{3x+3}}{5^{x-1}} = 5^{(3x+3)-(x-1)} = 5^{3x+3-x+1} = 5^{2x+4}$

6. ## Re: Year 10 Index Laws help please

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