PDA

View Full Version : Exponents and roots help

becomingagodo
2nd July 2007, 04:09 PM
Well, I reading through the Chapter 5 in Algebra demystified I came across this
6x^2 x^(1/2)=
You're mean't to rewrite it so it is simpler.
The anwser came out as 6x^5/2
Can someone explain this.
3/2 ^(1/2)= 6^(1/2)/2
Can someone also explain this.
16x^8^(1/4)= 2x^2
Seriously this must be wrong, how the hell does that equal 2x^2. Before I end up stabbing myself out off anger, someone please help me.

Sorry, I don't know how to express square roots or roots, so I had to do the power trick. I anwsered about 15/18 correct, however these three problems got me.

Ziggurat
2nd July 2007, 04:27 PM
Well, I reading through the Chapter 5 in Algebra demystified I came across this
6x^2 x^(1/2)=
You're mean't to rewrite it so it is simpler.
The anwser came out as 6x^5/2
Can someone explain this.
3/2 ^(1/2)= 6^(1/2)/2
Can someone also explain this.
16x^8^(1/4)= 2x^2
Seriously this must be wrong, how the hell does that equal 2x^2. Before I end up stabbing myself out off anger, someone please help me.

Sorry, I don't know how to express square roots or roots, so I had to do the power trick. I anwsered about 15/18 correct, however these three problems got me.

It actually simplifies things if you use exponentials rather than roots. It also helps sometimes to use parentheses to make the order of operation more explicit.

Your first problem: 6*(x^2) *(x^(1/2))
Well, 2 = 4/2, right? So x^2 = x^(4/2). Now when you multiply two exponentials with the same base (in this case, x), then you add the exponentials. For example, (x^3) * (x^2) = (x*x*x)*(x*x) = (x*x*x*x*x) = (x^5) = (x^(3+2)). In your case, though, the exponents are 1/2 and 2, so when you add them you get 5/2.

Second problem: exponential of a fraction is the fraction of the exponentials. (3/2)^(1/2) = (6/4)^(1/2) = (6^(1/2))/(4^(1/2)) = (6^(1/2))/2

Third problem: to take the exponential of an exponential, you multiply the powers. So for example, (x^3)^2 = (x*x*x)^2 = (x*x*x)*(x*x*x) = x^6 = x^(2*3). Furthermore, the exponential of a product of two numbers is the product of the exponentials: (y*x)^2 = (y*x)*(y*x) = (y*y)*(x*x) = (y^2)*(x^2). In your case, you have:
(16*(x^8))^(1/4) = (16^(1/4)) * ((x^8)^1/4) = 2 * x^2

Make sense?

Jekyll
2nd July 2007, 04:36 PM
Ok, so:
(6x^2)*( x^(1/2))= 6*(x^4/2)*( x^(1/2))
and we can add exponents when we multiply two expressions with the same bottom (x).
so:
(6x^2)*( x^(1/2))=6* (x^ ((4+1)/2))

Now, the next one is different. The rule is:
(a^b)^c = a^(b*c)
for all a, b and c.

Can you work it out from this?

And relax, maths is meant to be more fun than this, don't take it so seriously.

andyandy
2nd July 2007, 04:44 PM
Well, I reading through the Chapter 5 in Algebra demystified I came across this
6x^2 x^(1/2)=
You're mean't to rewrite it so it is simpler.
The anwser came out as 6x^5/2
Can someone explain this.
3/2 ^(1/2)= 6^(1/2)/2
Can someone also explain this.
16x^8^(1/4)= 2x^2
Seriously this must be wrong, how the hell does that equal 2x^2. Before I end up stabbing myself out off anger, someone please help me.

Sorry, I don't know how to express square roots or roots, so I had to do the power trick. I anwsered about 15/18 correct, however these three problems got me.

the first one

$6x^2.x^{0.5} just requires that you remember$ x^n.x^m = x^{n+m}

so

$6x^2.x^{0.5} = 6x^{2+0.5} = 6x^{2.5} = 6x^{5/2} The second one requires that you know$ (\frac{a}{b})^n = \frac{a^n}{b^n} $but is also a bit tricky.... first you need to write the fraction in a different form$ \frac{3}{2} = \frac{(3)(2)}{(2)(2)}

all you've done here is multiply both top and bottom by 2 - so the fraction is still the same - 3/2 = 6/4.

now you can apply the formula;

$(\frac{3}{2})^{0.5} = \frac{((3)(2))^{0.5}}{((2)(2))^{0.5}} = \frac{6^{0.5}}{2} The last one requires that you know$ (ax)^n = a^n.x^n

so first write your equation as

$(16x^8)^{0.25} = (16)^{0.25}.(x^8)^{0.25} = 2.x^{8(0.25)} = 2x^2 hope that helps.... :D i see that half of JREF has already answered in the time it took to type that in latex....oh well.... :) becomingagodo 2nd July 2007, 04:46 PM Thank you, I understand it now. 2.5 how do you turn this into a improper fraction? I forgot the rule again. andyandy 2nd July 2007, 05:36 PM Thank you, I understand it now. 2.5 how do you turn this into a improper fraction? I forgot the rule again. all you need to do is get 2.5 to being an integer (1,2,3 etc), so in this case all you need to do is multiply by 2....$ 2.5 = \frac{(2.5)(2)}{2} = \frac{5}{2}

More generally if you have any number with 1 decimal place, you can convert it to a fraction by multiplying by 10....

$2.5 = \frac{(2.5)(10)}{10} = \frac{25}{10} Once you have this, if the numerator and denominator have a common factor, you can simplify. In this case both 25 and 10 can be divided by 5, so$ \frac{25}{10} = \frac{5}{2}