Simplifying Exponents (I tried)

BilalAnsar

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(2x)-4 = (3x)-8

2-4 * x-4 = 3-8 * x-8

1/24 * 1/x4 = 1/38 * 1/x8

(1/24) / (1/38) = (1/x8) / (1/x4)

(1/24) * (38) = 1/x8 * x4

38 / 24 = x4-8

38 / 24 = x-4

38 / 24 = 1/x4

38 * x4 = 24

x4 = (24/38)

I don't know how to write it here but then I took fourth roots of both sides that gave me the answer 2/9 = 0.222...

However, the correct answer is supposed to be 0 or 0.75. I know there are other ways I could go about getting the right answer, but I really wish to understand what am I doing wrong above. Your help would be deeply, sincerely appreciated. Regards, a (GRE) Prepper.
 
(2x)-4 = (3x)-8

2-4 * x-4 = 3-8 * x-8

1/24 * 1/x4 = 1/38 * 1/x8

(1/24) / (1/38) = (1/x8) / (1/x4)

(1/24) * (38) = 1/x8 * x4

38 / 24 = x4-8

38 / 24 = x-4

38 / 24 = 1/x4

38 * x4 = 24

x4 = (24/38) ...since we are looking at the 4th root - there would be 3 other roots but none of those are 0 or 0.75.

I don't know how to write it here but then I took fourth roots of both sides that gave me the answer 2/9 = 0.222...

However, the correct answer is supposed to be 0 or 0.75. I know there are other ways I could go about getting the right answer, but I really wish to understand what am I doing wrong above. Your help would be deeply, sincerely appreciated. Regards, a (GRE) Prepper.
I'll do the problem slightly differentlY:

(2x)-4 = (3x)-8

2-4 * x-4 = 3-8 * x-8

[x-4] / [x-8] = [3-8] / [2-4]

x[-4 - (-8)] = [32](-4) / [2-4]

x[4] = [9/2](-4)

x[4] = [2/9]4

x = 2/9...............since we are looking at the 4th root - there would be 3 other roots but none of those are 0 or 0.75.

I get the same answer as you did. If you posted the "complete" problem correctly. You said:

"I know there are other ways I could go about getting the right answer," - can you please post one of those other ways?
 
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(2x)-4 = (3x)-8
...
I don't know how to write it here but then I took fourth roots of both sides that gave me the answer 2/9 = 0.222...

However, the correct answer is supposed to be 0 or 0.75. I know there are other ways I could go about getting the right answer, but I really wish to understand what am I doing wrong above. Your help would be deeply, sincerely appreciated. Regards, a (GRE) Prepper.
Did you try checking their answers? It should be clear that 0 can't be a solution; and 3/4 also is not, but 2/9 works.

Did you read the answer for the wrong problem? Nothing you did was wrong (though a lot could have been done a little easier.
 
Thank you very much guys. I showed my working from the middle and have realized my mistake now. The step I missed in the beginning was:

(3x)^1/8 = (2x)^1/4

This should simplify to
1/((3x)^-1/8) = 1/((2x)^-1/4)

What I did was the following:
(3x)^1/8 = (2x)^1/4
1/((3x)^-8) = 1/((2x)^-4)
(2x)^-4 = (3x)^-8

Out of curiosity though, how do I find the other three roots?
 
Thank you very much guys. I showed my working from the middle and have realized my mistake now. The step I missed in the beginning was:

(3x)^1/8 = (2x)^1/4

This should simplify to
1/((3x)^-1/8) = 1/((2x)^-1/4)

What I did was the following:
(3x)^1/8 = (2x)^1/4
1/((3x)^-8) = 1/((2x)^-4)
(2x)^-4 = (3x)^-8

Out of curiosity though, how do I find the other three roots?
x4 - a4 = 0

(x2 - a2)(x2 + a2) = 0

(x - a)(x + a)(x + ia)(x - ia) = 0
 
Thank you very much guys. I showed my working from the middle and have realized my mistake now. The step I missed in the beginning was:

(3x)^1/8 = (2x)^1/4

This should simplify to
1/((3x)^-1/8) = 1/((2x)^-1/4)

What I did was the following:
(3x)^1/8 = (2x)^1/4
1/((3x)^-8) = 1/((2x)^-4)
(2x)^-4 = (3x)^-8

Out of curiosity though, how do I find the other three roots?
Never, ever ask for help without showing the original problem. A lot of time is wasted by people who assume their work so far is perfect.

There is no need to use negative exponents here at all. And I presume you are now aware that a^{1/n} does not equal either a^{-n} or 1/a^{-n}, as you suppose in your work.

To solve the actual problem, (3x)^{1/8} = (2x)^{1/4}, I would raise each side to the 8th power, leading to 3x = 4x^2. Then I'd solve that equation. I'd also want to make sure that raising to an even power didn't change the solutions (e.g. resulting in an extraneous solution).

The mention of "the other three roots" applied to the wrong equation you gave us.
 
Here is how I would have solved this problem:

(2x)-4 = (3x)-8

Step 1: Take the reciprocal of both sides! (2x)4 = (3x)8. I got rid of all the negative exponents.
Now continue from this point as others have shown.
 
Here is how I would have solved this problem:

(2x)-4 = (3x)-8

Step 1: Take the reciprocal of both sides! (2x)4 = (3x)8. I got rid of all the negative exponents.

A warning would be that these are not equivalent equations, because x = 0 would
be a solution to the latter. However, it would not be a solution to the former, because zero cannot be raised to an odd exponent.
 
A warning would be that these are not equivalent equations, because x = 0 would
be a solution to the latter. However, it would not be a solution to the former, because zero cannot be raised to an odd exponent.
I think you mean to a negative exponent.

In other words, taking the reciprocal of both sides doesn't work if the values are zero.

And, of course, this isn't the real problem being solved, according to post #4.
 
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