Question 277174
{{{A=6V^(2/3)}}}
You're given A so:
{{{12 = 6V^(2/3)}}}
We're solving for V so we need to get rid of the 6 in front and the exponent of 2/3. We can eliminate the 6 by dividing both sides by 6:
{{{2 = V^(2/3)}}}
Now how do we get rid of the exponent? Well, to be perfectly correct, we are not getting rid of the exponent we are changing it to a 1 (since {{{V^1 = V}}}. So how do we change the exponent to a 1?<br>
One way is to raise each side to the 3/2 power. (If you don't see why this works yet, you will shortly.)
{{{(2)^(3/2) = (V^(2/3))^(3/2)}}}
On the right side, the rule is to multiply the exponents. 2/3 and 3/2 are reciprocals of each other. And what do you get when you multiply reciprocals? Answer 1! This is why I raised both sides to the 3/2 power:<ul><li>I knew that raising a power to a power would result in multiplying exponents</li><li>I knew that I wanted an exponent of 1</li><li>I knew that multiplying reciprocals always results in 1</li><li>3/2 is the reciprocal of 2/3</li></ul>
{{{(2)^(3/2) = V}}}
Now we only have to simplify the left side. If you are not yet comfortable with fractional exponents, it can be helpful to look at the exponent in factored form. {{{3/2 = 3*(1/2)}}} The factor of three in the exponent says we'll be cubing something and the factor of 1/2 in the exponent says there will be a square root. And, since multiplication is Commutative, we can do these things in any order we choose. Since 2 is not a prefect square, doing the square root first has little appeal. So I'll start by cubing 2. 2 cubed is 8 and then we apply the square root giving us {{{sqrt(8)}}}. So:
{{{V = 2^(3/2) = sqrt(8)}}}
The only thing left is to simplify the square root. There is a perfect square factor in 8:
{{{V = 2^(3/2) = sqrt(8) = sqrt(4*2) = sqrt(4)*sqrt(2) = 2sqrt(2)}}}