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adr8
08-02-2011, 05:12 PM
I got the answers for these problems. I just dont know if I got them right. They told me to just solve each equation and to round x to four significant digits if necessary. Anyways, here are my problems and how I solved them:

1 (x/y)^1/3 (y/x)^2/3
Step 1. x(^1/3) y(^2/3)/ x(^2/3) y(^1/3)

Step 2. x^-1/3 y^1/3

Answer: y(^1/3)/x(^1/3)

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 I turned the (x-4)^1/3 in to a radical so its 2 times the radical of (x-4)^1/3.

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.

Step2. So far I have 4* 1/(x-4)^2/3.

The answer that I have is 4/(x-4)^2/3. Then I changed it to the radical form in which I know the exponent 2 would be a negative.

JeffM
08-02-2011, 05:50 PM
I got the answers for these problems. I just dont know if I got them right. They told me to just solve each equation and to round x to four significant digits if necessary. Anyways, here are my problems and how I solved them:

1 (x/y)^1/3 (y/x)^2/3 This is very hard to interpret. I presume you mean [(x/y)[sup:3tk8trde]1/3[/sup:3tk8trde]] * [(y/x)[sup:3tk8trde]2/3[/sup:3tk8trde]]. That, however, is not an equation; it is an expression. It cannot be SOLVED. If the expression is correct, it does indeed simplify to

Step 1. x(^1/3) y(^2/3)/ [x(^2/3) y(^1/3)] The brackets are NECESSARY. That expression does simplify to

Step 2. x^-1/3 * y^1/3 At a minimum you need the * in there. That expression does simplify to

Answer: y(^1/3)/x(^1/3) But you have not rounded anything to four significant digits. Part of the problem seems to be missing.

2. (x-4)^-1/3=2 THIS IS AN EQUATION.

Step 1. 1/ (x-4)^1/3=2 Yes

Step 2. I moved the (x-4)^1/3 to the other side so my answer is 2(x-4)^1/3 EQUALS WHAT? You cannot just ignore one half of an equation. In any case, moving the power to the other side does result in 2(x - 4)[sup:3tk8trde](1/3)[/sup:3tk8trde] = 1.
I turned the (x-4)^1/3 in to a radical so its 2 times the radical of (x-4)^1/3. No, it is 2 times the cube root of (x - 4), not the cube root of (x - 4)(1/3). In any case, using a radical sign does not help you solve the equation. The trick is to cube each side of the equation. You take it from here.

3. (x-4)^2/3=4 Look at what I said about problem 2 and try this one again.

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.

Step2. So far I have 4* 1/(x-4)^2/3.

The answer that I have is 4/(x-4)^2/3. Then I changed it to the radical form in which I know the exponent 2 would be a negative.

adr8
08-02-2011, 07:03 PM
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?

2. (x-4)^-1/3=2 THIS IS AN EQUATION.

Step 1. 1/ (x-4)^1/3=2 Yes

Step 2. I moved the (x-4)^1/3 to the other side so my answer is 2(x-4)^1/3 EQUALS WHAT? You cannot just ignore one half of an equation. In any case, moving the power to the other side does result in 2(x - 4)(1/3) = 1. I did think about equalling it to 1, but I wasnít sure when we cancelled a fraction if the 1 was still there or it would disappear since it got cancelled, but now it makes perfect sense.

I turned the (x-4)^1/3 in to a radical so its 2 times the radical of (x-4)^1/3. No, it is 2 times the cube root of (x - 4), not the cube root of (x - 4)(1/3). In any case, using a radical sign does not help you solve the equation. The trick is to cube each side of the equation. You take it from here. Yea that is what I meant cube root sorry I just didnít couldnít remember how it would be called when you have the number 3 outside the square root as a small number lol.
Anyways I ended up cube rooting the 1 so it would cancel the cube root of (x-4).
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.
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.

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.
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.
Step 5. Then I negative square rooted (x-4).
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

Denis
08-02-2011, 09:45 PM
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.

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.
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.
Step 5. Then I negative square rooted (x-4).
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
Yikes! What are you doing? :shock:
(x - 4)^(2/3) = 4
x - 4 = 4^(3/2) ......RULE: if a^x = b, then a = b^(1/x)
x - 4 = 4^1 * 4^(1/2) .......RULE: a^x * a^y = a^(x+y)
x - 4 = 4 * 2
x - 4 = 8
x = 12

adr8
08-03-2011, 02:17 AM
3. (x-4)^2/3=4


Yikes! What are you doing? :shock:
(x - 4)^(2/3) = 4
x - 4 = 4^(3/2) ......RULE: if a^x = b, then a = b^(1/x)
x - 4 = 4^1 * 4^(1/2) .......RULE: a^x * a^y = a^(x+y)
x - 4 = 4 * 2
x - 4 = 8
x = 12

Oh okay, the only question I have is on the third or second step why do you have to put another four?

By any chance did I get the 2nd question right? My answer is 4.5. Its on my last post.

You know what I was noticing that I forgot to put the 1/3 as a negative. But my answer is still 4.5. Sorry for the typo....

JeffM
08-03-2011, 12:38 PM
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:3230nb74](1/3)[/sup:3230nb74]][sup:3230nb74](1/3)[/sup:3230nb74]. 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:3230nb74](1/3)[/sup:3230nb74] = 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:3230nb74](-3/5)[/sup:3230nb74] = 2
Step 1: Get rid of the negative exponent (if any) by using the reciprocal
54 / (x - 2)[sup:3230nb74](3/5)[/sup:3230nb74] = 2
Step 2: Get rid of the fraction (if any and if possible) by cross multiplying.
(x - 2)[sup:3230nb74](3/5)[/sup:3230nb74] = 54/2 = 27.
Step 3: Get rid of the NUMERATOR in the fractional exponent by taking a ROOT.
[(x - 3)[sup:3230nb74](3/5)[/sup:3230nb74]][sup:3230nb74](1/3)[/sup:3230nb74] = 27[sup:3230nb74](1/3)[/sup:3230nb74].
(x - 3)[sup:3230nb74](1/5)[/sup:3230nb74] = 3
Step 4: Get rid of the DENOMINATOR in the fractional exponent by taking a POWER.
[(x - 3)[sup:3230nb74](1/5)[/sup:3230nb74]][sup:3230nb74]5[/sup:3230nb74] = 3[sup:3230nb74]5[/sup:3230nb74]
x - 3 = 247.
Step 4: Gather like terms.
x - 3 + 3 = 247 + 3
x = 250.

adr8
08-03-2011, 06:39 PM
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.


I am confused on the example. I got lost in step 3. Why is it 1/3 and not 1/5? I kind of see why its 1/3, but not completely. Also, why is it x-3 shouldnt it be x-2?

Denis
08-03-2011, 11:24 PM
Let's try an example.
54 * (x - 2)[sup:14dfesrk](-3/5)[/sup:14dfesrk] = 2
Step 1: Get rid of the negative exponent (if any) by using the reciprocal
54 / (x - 2)[sup:14dfesrk](3/5)[/sup:14dfesrk] = 2
Step 2: Get rid of the fraction (if any and if possible) by cross multiplying.
(x - 2)[sup:14dfesrk](3/5)[/sup:14dfesrk] = 54/2 = 27.
Step 3: Get rid of the NUMERATOR in the fractional exponent by taking a ROOT.
[(x - 3)[sup:14dfesrk](3/5)[/sup:14dfesrk]][sup:14dfesrk](1/3)[/sup:14dfesrk] = 27[sup:14dfesrk](1/3)[/sup:14dfesrk].
(x - 3)[sup:14dfesrk](1/5)[/sup:14dfesrk] = 3
Step 4: Get rid of the DENOMINATOR in the fractional exponent by taking a POWER.
[(x - 3)[sup:14dfesrk](1/5)[/sup:14dfesrk]][sup:14dfesrk]5[/sup:14dfesrk] = 3[sup:14dfesrk]5[/sup:14dfesrk]
x - 3 = 247.
Step 4: Gather like terms.
x - 3 + 3 = 247 + 3
x = 250.
I am confused on the example. I got lost in step 3. Why is it 1/3 and not 1/5? I kind of see why its 1/3, but not completely. Also, why is it x-3 shouldnt it be x-2?

(x - 2)^(3/5) = 27
Go this way:
x - 2 = 27^(5/3)
x - 2 = 243
x = 245

adr8
08-04-2011, 02:51 PM
Thank you so much :), its more clearer now. To be on the safe side I was wondering if you can come up with two of these problems to see if I can solve them. I ran out of those problems and I want to see if I can solve them. Thanks :).

adr8
08-04-2011, 05:53 PM
My answer to the first one is -11 and the answer to the second one is 33.

8 * (36 + x)(3/2) = 1000

Step 1. [8 * (36 + x)(3/2)(2/3)] = 1000(2/3) in other words, i cancelled the exponent from one side and put it on the otherside by using 2/3.

Step 2. 4(36+x)=100 This is what I ended up with. Then I multiplied four by 36 and by x.

Step 3. 144+4x=100. I ended up with this. I moved the 144 to the otherside by cancelling it out by -144.

Step 4. 4x=-44. I ended up wiht this and my answer came out to -11.


20 * (x - 1)-(2/5) = 5

Step 1. 20 * 1/(x - 1)(2/5) = 5.

Step 2. 20/(x-1)(2/5)=5

Step 3. 20=5(x-1)(2/5)

Step 4. 205/2=[5(x-1)2/5]5/2 I ended up cancelling the exponent by 5/2 in order to move it to the other side.

Step 5. 1788.8544=55.9017(x-1)

Step 6. 1788.8544=55.9017x- 55.9017

Step 7. 1844.7561=55.9017x I moved 5590.17 to the other side.

Step 8 I ended up dividing them and I got 33=x.

adr8
08-05-2011, 02:42 PM
Yep and thank u and Dennis so much for the help :). I will write down your method as well that way I can study it more.