sqrt(2x) = 10
Looking for x
I can't figure out how to do this (obviously I know the answer is 50, since 2x50 =100 and the sqrt(100) is 10) but I can't solve it.
Thanks for your help.
$\displaystyle \sqrt{2x} = 10$
Let's remember some exponents rules:
$\displaystyle (a^n)^m = a^{nm}$
$\displaystyle \sqrt[n]{a} = a^{\frac{1}{n}} $
$\displaystyle (a^{\frac{1}{n}})^n = a^{\frac{n}{n}} = a^1 = a$
Also, remember the principal of powers:
If a = b, then $\displaystyle a^n = b^n.$
Let's go back to the original question:
$\displaystyle \sqrt{2x} = 10$
Don't you agree it's actually:
$\displaystyle (2x)^{\frac{1}{2}} = 10$
But hey, I wanna get rid of this fractional exponent, so we can isolate x. We do that by squaring both sides.
$\displaystyle ((2x)^{\frac{1}{2}})^2 = (10)^2$
2x = 100
Now solving for x becomes a simple matter.
$\displaystyle 0.5^2$ does not mean 2 multiplied by 0.5. 2 is an exponent. What it means is that you have to multiply 0.5 twice. For instance:
$\displaystyle 0.5^2 = 0.5 \times 0.5 = 0.25 = \frac{1}{4}$
$\displaystyle 3^2 = 3 \times 3 = 9$
$\displaystyle 5^4 = 5 \times 5 \times 5 \times 5 = 625$
$\displaystyle 4^3 = 4 \times 4 \times 4 = 64$
To generalize this:
$\displaystyle a^n = \underbrace{a \times a \times a \times \cdots \times a}_{\text{n\ times}}$
When 2 is written in superscript of a number, then it is an exponent.
More examples:
$\displaystyle (5^2)^3 = 5^6 = 5 \times 5 \times 5 \times 5 \times 5 \times 5$
$\displaystyle (6^1)^2 = 6^2 = 6 \times 6 = 36$
$\displaystyle (7^3)^3 = 7^9 = 7 \times 7 \times 7 \times 7 \times 7 \times 7 \times 7 \times 7 \times 7 = 40353607$
OK, so when I do $\displaystyle (2x^\frac{1}{2})^2$ what I'm ask to do is take 0.5 x 2, which is 1, and then put $\displaystyle (2x)^1$ which is 2x.Originally Posted by Chop Suey;175491
When 2 is written in superscript of a number, then it is an exponent.
More examples:
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And that's how I get 2x = 100 or x=50.
OK, I don't "get it" but I understand what you are saying. Which might be as far as I can get this.
These are the properties of exponents. Let me try to explain it to you again. Let $\displaystyle a$ denote a nonzero real number.
$\displaystyle a = a^1$
$\displaystyle a \times a = a^2$
$\displaystyle a \times a \times a \times a = a^4$
$\displaystyle a \times a \times a \times a \times a \times a \times a \times a \times a \times a = a^{10}$
Let's reverse this:
$\displaystyle a^1 = a$
$\displaystyle a^2 = a \times a$
$\displaystyle a^4 = a \times a \times a \times a$
$\displaystyle a^{10} = a \times a \times a \times a \times a \times a \times a \times a \times a \times a$
See what we are doing here? We are condensing this long expression, and we do that by writing a number in superscript near the number to indicate how many times it is multiplied. To generalize this, let's say:
$\displaystyle a^n = \underbrace{a \times a \times a \times a \times \cdots \times a}_{\text{n times}}$
$\displaystyle \underbrace{a \times a \times a \times a \times \cdots \times a}_{\text{n times}} = a^n$
============
Now, we got that covered. Now, what if we multiply two powers?
$\displaystyle a^4 \times a^5$
Think about it. If you expand it:
$\displaystyle \underbrace{a \times a \times \cdots \times a}_{\text{4+5 times}}$
Don't you agree that:
$\displaystyle a^4 \times a^5 = a^{4+5} = a^9$
To generalize this, we say:
$\displaystyle a^n \times a^m = a^{n+m}$
============
Moving on to the power of a power. Suppose you have:
$\displaystyle (a^4)^5$
Let's expand now:
$\displaystyle a^4 \times a^4 \times a^4 \times a^4 \times a^4$
Now, I told you what happens when we multiply two powers. This expression above becomes:
$\displaystyle a^{4+4+4+4+4}$
Notice that 4 appears 5 times. Thus:
$\displaystyle = a^{4\cdot 5} = a^{20}$
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Moving on...what happens when I take the quotient of powers? Well, let me introduce to you the definition of exponents first:
Assume a and n are real numbers. Then:
$\displaystyle a^n = \underbrace{a \times a \times a \times a \times \cdots \times a}_{\text{n times}}$
$\displaystyle a^0 = 1$
$\displaystyle a^{-n} = \frac{1}{a^n}$
Going back to the original problem: quotient of powers.
$\displaystyle \frac{a^n}{a^m} = a^n \times \frac{1}{a^m}$
Don't you agree that:
$\displaystyle \frac{1}{a^m} = a^{-m}$
And thus:
$\displaystyle a^n \times a^{-m} = a^{n+(-m)} = a^{n-m}$
$\displaystyle \frac{a^n}{a^m} = a^{n-m}$
============
Now, dealing with difference powers. It should seem obvious to you now that
$\displaystyle (ab)^n = a^nb^n$
$\displaystyle \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$
If it's easier, think about what's happened to the x in order to turn it into 10.
It's been multiplied by 2, then square rooted.
So, by working in reverse and using inverse operations we can find x.
The first thing we undo is the square rooting, by squaring both sides of the equation.
So
$\displaystyle \sqrt{2x}=10$
$\displaystyle (\sqrt{2x})^2=10^2$
$\displaystyle 2x=100$
Now we undo the multiplying by 2, by dividing both sides by 2.
$\displaystyle \frac{2x}{2}=\frac{100}{2}$
$\displaystyle x=50$
Ok, thank you Chop Suey, for taking the time to go through this. I get all that you explained.
Thank you Prove it, it's better for me to do this particular problem the way you did it.
I'll go back over everything Monday, and make sure I didn't mess up somewhere in the homework.
I'm almost done!