Help Solving Equation: 0,5m^4/5 = 2u^1/5

  • Thread starter storoi1990
  • Start date
Anyway, I think it is good for our summary purposes.In summary, the conversation was about a problem with the equation 0,5m^4/5 = 2u^1/5 where the variables u and m were supposed to be under the fractions. The solution involved cross-multiplying and simplifying the equation to get m = 4u. However, there was a small error in the solution where the exponent should have been put in parentheses.
  • #1
storoi1990
14
0

Homework Statement



I need help with this euation:

0,5m^4/5 = 2u^1/5
u^4/5 m^1/5

it's been calculated to be:

0,5m = 2u => m = 4u

i need the calculations inbetween..

So far i have come to think that they cross out 4/5, on one side.

and on the other they have crossed out 1/5.

and they are left with:

0,5m = 2u
u m


But what now?


Note: (u) is supposed to be under 0,5m, and (m) is supposed to be under 2u

Likewise with the one on top of the page.

Homework Equations





The Attempt at a Solution

 
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  • #2
Your equation seems to be
[tex]\frac{.5m^{4/5}}{u^{4/5}} = \frac{2u^{1/5}}{m^{1/5}}[/tex]

An obvious thing to do would be to multiply both sides by u4/5m1/5. This is equivalent to what is called cross-multiplying. Don't think of this as "crossing out" anything.

From the equation above, you get .5m = 2u. Can you take it from there?
 
  • #3
Your equation seems to be
LaTeX Code: \\frac{.5m^{4/5}}{u^{4/5}} = \\frac{2u^{1/5}}{m^{1/5}}

An obvious thing to do would be to multiply both sides by u4/5m1/5. This is equivalent to what is called cross-multiplying. Don't think of this as "crossing out" anything.

From the equation above, you get .5m = 2u. Can you take it from there?

yeah, you have typed question wrong. or otherwise there is a misprinting in your book either in question or in solution.

See the answer for your given question
[tex]
\frac{0.5m^{4/5}}{u^{4/5}.m^{1/5}} = 2u^{1/5}
[/tex]

[tex]
\frac{m^{4/5}}{u^{4/5}.m^{1/5}} = 4u^{1/5}
[/tex]

[tex]

m^{3/5} = 4u

[/tex]

[tex]

m = (4u)^{5/3}

[/tex]
 
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  • #4
to snshusat:

yes, it's supposed to be the way mark wrote.

how do you write it like that btw?

to mark:

okey so then i get 0,5m^4/5*m^1/5 = 2u^1/5*u^4/5

so: 0,5m^5/5 = 2u^5/5

and then: 0,5m = 2u

and then finally: m = 4u

right?
 
Last edited:
  • #6
snshusat161 said:
[tex]

m^{3/5} = 4u

[/tex]

[tex]

m = {4u}^{5/3}

[/tex]

Even though the question was different, this is incorrect.

[tex]m^{3/5}=4u[/tex]

[tex]m=(4u)^{5/3}=4^{5/3}u^{5/3}[/tex]

edit: after seeing your latex as {4u}^{5/3} you weren't wrong, but you should have added parentheses around the (4u).
 
  • #7
thanks for figuring it out, this code is very hard to type.

Edit: I think I should have used () in place on {}. I thought that I've become so expert in typing those codes that I've not reviewed it again after posting. That guy may have got punishment from his teacher if he had copied that solution. Sorry :(
 
Last edited:
  • #8
snshusat161 said:
That guy may have got punishment from his teacher if he had copied that solution. Sorry :(

Sure he would have! You started with the wrong question :-p
 

FAQ: Help Solving Equation: 0,5m^4/5 = 2u^1/5

How do I solve this equation?

To solve this equation, you can follow these steps:
1. Multiply both sides by 5 to get rid of the denominator.
2. Raise both sides to the power of 5 to eliminate the fractional exponents.
3. Simplify the equation and solve for the variable.

What is the value of the variable in this equation?

The value of the variable can be found by solving the equation using the steps mentioned above.

Can this equation be solved without a calculator?

Yes, this equation can be solved without a calculator by following the steps mentioned in the first question.

What is the significance of the numbers in this equation?

The numbers in this equation represent the coefficients and exponents of the variables. They are used to manipulate and solve the equation.

Is there a faster way to solve this equation?

There may be alternative methods to solve this equation, such as using logarithms or substitution, but it ultimately depends on the individual and their level of understanding in mathematics.

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