adr8 said:
Oh ok, I think I got it:
For the first problem my bad I did relook at what the problem was asking me to do and yea you are correct it is an expression. The instructions were for the other two problems not this one. This one just asked me to simplify the expression. For some weird reason I thought they were together lol. Oh well, so I did get that one right? Yes
2. (x-4)^-1/3=2
Step 1. 1/ (x-4)^1/3=2
Step 2. I moved the (x-4)^1/3 to the other side so my answer is 2(x-4)^1/3 = 1.
Anyways I ended up cube rooting the 1 so it would cancel the cube root of (x-4).
Here is where your problem lies. If you take the cube root of 1, you have to take the cube root of 2(x - 4)(1/3) = [2(x - 4)[sup:2sv3qxoe](1/3)[/sup:2sv3qxoe]][sup:2sv3qxoe](1/3)[/sup:2sv3qxoe]. When you change one side of an equation you must change the other side IN THE EXACT SAME WAY, right?
So, as I said in my previous post, to move toward x you need to CUBE BOTH sides of the equation 2(x - 4)[sup:2sv3qxoe](1/3)[/sup:2sv3qxoe] = 1, which gives you
8(x - 4) = 1.
Afterwards, I ended up multiplying 2(x-4) in which my answer was 2x-8=1.
I cancelled negative 8 by turning it to a positive. Then I moved +8 to the other side in which it equalled to 9. So far my equation is 2x=9 and my answer is 4.5 =x. This is wrong for the reason given above.
3. (x-4)^2/3=4
Step 1. I got the reciprocal of (x-4)^2/3 so my reciprocal would be 1/(x-4)^2/3 and I moved it to the other side. You do not need a reciprocal because there is no negative exponent. Consequently, moving to the other side just complicates things.
Step2. So far I have 4* 1/(x-4)^2/3.
Step 3. I got 4/(x-4)^2/3=1. I ended up changing it in to a radical form in which its 4 cube root (x-4)^-2. Where did the minus 2 come from? Changing from exponential form into radical form does not simplify the problem; it just changes notation into a less useful form.
Step 4. I cube rooted the 1 so it can cancel with the cube root of (x-4)^-2. I am left with 4(x-4)^-2=1. This is the same error as you made in the previous problem. You must do the same thing to both sides of the equation.
Step 5. Then I negative square rooted (x-4). Same error repeated, and what is negative square rooting.
Step 6. I cancelled the negative square root by negative square rooting the 1 and it gave me -1. So far I have -1=4(x-4).
Step 7. I am left with -1=4x-16. I moved -16 to the other side by cancelling it with a positive 16.
Step 8 I so far have 4x=15. My answer is x=3.750 That is not even close to what I get.
Let's try an example.
54 * (x - 2)[sup:2sv3qxoe](-3/5)[/sup:2sv3qxoe] = 2
Step 1: Get rid of the negative exponent (if any) by using the reciprocal
54 / (x - 2)[sup:2sv3qxoe](3/5)[/sup:2sv3qxoe] = 2
Step 2: Get rid of the fraction (if any and if possible) by cross multiplying.
(x - 2)[sup:2sv3qxoe](3/5)[/sup:2sv3qxoe] = 54/2 = 27.
Step 3: Get rid of the NUMERATOR in the fractional exponent by taking a ROOT.
[(x - 3)[sup:2sv3qxoe](3/5)[/sup:2sv3qxoe]][sup:2sv3qxoe](1/3)[/sup:2sv3qxoe] = 27[sup:2sv3qxoe](1/3)[/sup:2sv3qxoe].
(x - 3)[sup:2sv3qxoe](1/5)[/sup:2sv3qxoe] = 3
Step 4: Get rid of the DENOMINATOR in the fractional exponent by taking a POWER.
[(x - 3)[sup:2sv3qxoe](1/5)[/sup:2sv3qxoe]][sup:2sv3qxoe]5[/sup:2sv3qxoe] = 3[sup:2sv3qxoe]5[/sup:2sv3qxoe]
x - 3 = 247.
Step 4: Gather like terms.
x - 3 + 3 = 247 + 3
x = 250.