Indices using algebra

I'm really confused on how to solve for x in an equation that involves indices.

For example, a question I have is x^3/2 = 2x^2. I currently have (√x)^3 = 2x^2 and I am unsure of where to go from here.

According to the textbook the answer is 1/4, but I really can't see how to achieve this.

Any help would be greatly appreciated, thank you!

Reply 1

HandmadeTurnip

17

~~Try dividing everything by $x^2$ .~~

EDIT: Sorry, ignore that. While that would give you one answer, you'd miss out on the other possible answer.

(edited 9 years ago)

Reply 2

Ninja Saru

x^3/2 = 2x^2
(Now square everything)
x^3 = 4x^4
(divide both sides by x^3)
1 = 4x
Therefore, x = 1/4

Original post by HandmadeTurnip

Try dividing everything by

x^2

.

Dismal advice. You just blew up the universe.

Original post by Ninja Saru

x^3/2 = 2x^2
(Now square everything)
x^3 = 4x^4
(divide both sides by x^3)
1 = 4x
Therefore, x = 1/4

The fabric of spacetime is now destroyed for a second time. Well done.

Reply 5

ellie3xo

OP

Original post by Ninja Saru

x^3/2 = 2x^2
(Now square everything)
x^3 = 4x^4
(divide both sides by x^3)
1 = 4x
Therefore, x = 1/4

Thank you so much, this makes complete sense! I guess I was going off on a tangent trying to do something different, must have got confused in the way we were taught. Thank you again!

Original post by ellie3xo

...

Subtract

2x^2

from each side to obtain

x^{\frac{3}{2}}-2x^2 = 0

Now factorise and solve.

Reply 7

Ninja Saru

Original post by Mr M

The fabric of spacetime is now destroyed for a second time. Well done.

Wait, am I wrong...... I hope not.......

Reply 8

Ninja Saru

Original post by ellie3xo

Thank you so much, this makes complete sense! I guess I was going off on a tangent trying to do something different, must have got confused in the way we were taught. Thank you again!

Glad I could help

Original post by Ninja Saru

Wait, am I wrong...... I hope not.......

Reply 10

Ninja Saru

Original post by Mr M

huh?

Original post by ellie3xo

Thank you so much, this makes complete sense! I guess I was going off on a tangent trying to do something different, must have got confused in the way we were taught. Thank you again!

The answer you were given is incomplete and the method is wrong.

Reply 12

ellie3xo

OP

Original post by Mr M

The answer you were given is incomplete and the method is wrong.

The answer in the textbook is 1/4 and the method is way closer to what we were taught in class than the method in which you gave me, which doesn't even use indice rules.

Original post by Ninja Saru

huh?

If you were asked to solve

y^2=5y

for example I hope you would not simply divide by y. Did that help?

Original post by ellie3xo

The answer in the textbook is 1/4 and the method is way closer to what we were taught in class than the method in which you gave me, which doesn't even use indice rules.

Ok then the textbook is wrong unless you have not posted the complete question.

The method I gave you certainly uses the laws of indices.

Reply 15

ellie3xo

OP

The complete question in the textbook is to solve the equation for x:

x^3/2 = 2x^2

The answer displayed in the back of the textbook reads '1/4'

What do you think the answer is by your method?

Original post by ellie3xo

What do you think the answer is by your method?

I hope that was not intended to be as rude as it sounded.

x = 0 is a solution. Try substituting this value into the equation on your calculator. When you were advised to divide by x squared or x cubed you were dividing by zero. This is not permitted in maths. Try dividing something by zero on your calculator.

Reply 17

ztibor

10

Original post by ellie3xo

I'm really confused on how to solve for x in an equation that involves indices.

For example, a question I have is x^3/2 = 2x^2. I currently have (√x)^3 = 2x^2 and I am unsure of where to go from here.

According to the textbook the answer is 1/4, but I really can't see how to achieve this.

Any help would be greatly appreciated, thank you!

For opeartion of powers with positive base of x with real exponent the following identites hold

\displaystyle x^{\alpha}\cdot x^{\beta}=x^{\alpha+\beta}

\displaystyle \frac{x^{\alpha}}{ x^{\beta}}=x^{\alpha-\beta}

\displaystyle \left (x^{\alpha}\right )^{\beta}=x^{\alpha \cdot \beta}

\displaystyle x^{\alpha}\cdot y^{\alpha}=\left (x\cdot y\right )^{\alpha}

\displaystyle \frac{x^{\alpha}}{ y^{\alpha}}=\left (\frac{x}{y}\right )^{\alpha}

So e.g.

\displaystyle \frac{x^{\frac{3}{2}}}{x^{\frac{2}{3}}}=x^{\frac{3}{2}-\frac{2}{3}}=x^{\frac{9}{6}-\frac{4}{6}}=x^{\frac{5}{6}}

(edited 9 years ago)

Reply 18

Ninja Saru

Actually yes, Mr M is right
I was thinking on the basis that the answer is only 1/4 so I based my method on that however this isn't the case
x^3/2 - 2x^2 =0
(Put x^3/2 outside of the brackets)
x^3/2 (1 - 2x^1/2) = 0
Therefore x^3/2 =0 and 1 - 2x^1/2 = 0
If x^3/2 = 0, x can only be 0... and if 1 - 2x^1/2 = 0, 2x^1/2 = 1
So x^1/2 = 1/2, so x = 1/4
THEREFORE X=0 or 1/4
Thanks to Mr M for pointing out my mistake :biggrin:

Reply 19

ellie3xo

OP

Original post by Mr M

I hope that was not intended to be as rude as it sounded.

x = 0 is a solution. Try substituting this value into the equation on your calculator. When you were advised to divide by x squared or x cubed you were dividing by zero. This is not permitted in maths. Try dividing something by zero on your calculator.

It wasn't rude, I simply asked what you thought the answer was because obviously the text book doesn't agree. Please may you fully explain the question using your method because I don't understand how to do it.

Indices using algebra

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