# Solve for X

• Nov 18th 2009, 08:14 PM
hovermet
Solve for X
Help, Solve for X
4x^-2/3 = 16

i got to the point where 1/3\sqrt{4x^2} = 16 , then i am totally lost (Sadsmile)

Also what is the rule for a base with a power of negative fraction? Why do you have to flip ex: x^-1/2. 1/\sqrt{x}

Thank You

PS: am i posting in the corrct subforum？。。.
• Nov 18th 2009, 08:22 PM
pickslides
I believe this is the correct sub forum

Can you please confirm the problem?

$\displaystyle 4x^{-2/3} = 16$

or

$\displaystyle (4x)^{-2/3} = 16$
• Nov 18th 2009, 08:43 PM
hovermet
Quote:

Originally Posted by pickslides
I believe this is the correct sub forum

Can you please confirm the problem?

$\displaystyle 4x^{-2/3} = 16$

or

$\displaystyle (4x)^{-2/3} = 16$

It is the top one $\displaystyle 4x^{-2/3} = 16$ , but do you mind demonstrating the bottom one also? ty
• Nov 18th 2009, 08:53 PM
Billyboy
Use the properties of exponentiation.

solving both ways...

$\displaystyle (4x)^{-2/3} = 16$

$\displaystyle (4x) = 16^{-3/2}$

$\displaystyle x = (16^{-3/2})/4$

$\displaystyle 4x^{-2/3} = 16$

$\displaystyle x = (16/4)^{-3/2}$
• Nov 18th 2009, 09:04 PM
hovermet
Quote:

Originally Posted by Billyboy
Use the properties of exponentiation.

solving both ways...

$\displaystyle (4x)^{-2/3} = 16$

$\displaystyle (4x) = 16^{-3/2}$

$\displaystyle x = (16^{-3/2})/4$

$\displaystyle 4x^{-2/3} = 16$

$\displaystyle x = (16/4)^{-3/2}$

Thank You Billy, the answer key for 4x^{-2/3} = 16 is 1/8, how do you simplify it into 1/8?.. i am having big problems with negative fraction powers(Bow)
• Nov 18th 2009, 09:24 PM
Billyboy
$\displaystyle x = (16/4)^{-3/2}$

$\displaystyle x = 4^{-3/2}$

raising a nonzero number to a "−" power produces its reciprocal...

$\displaystyle x^{-1} = 1/x$

and

$\displaystyle x^{-2}=1/x^2$

i.e.

$\displaystyle x^{-a}=1/x^a$

think you get the idea...

...taking what you know from above and understanding that the numerator raises the number to that power (in this case ^3) and the denominator takes the root (in this case 2nd root or sqrt) therefore you can simplify...

...This

$\displaystyle x = 4^{-3/2}$

is the same as this...

$\displaystyle x = 1/(4^{3/2})$

is the same as this...

$\displaystyle x = 1/sqrt(4^3)$

Therefore, we easily solve...

$\displaystyle 4^3 = 64$

and

$\displaystyle sqrt(64) = 8$

Thus you get...

$\displaystyle x = 1/8$
• Nov 18th 2009, 10:18 PM
Billyboy
a good property to know...

$\displaystyle (a/b)^{-1} = b/a$

Also, take for example:

$\displaystyle x^{a}=y$

to solve for x...

$\displaystyle x^{a(a^{-1})}=y^{(a^{-1})}$

$\displaystyle (a*a^{-1} = a * 1/a = 1)$

sooo...

$\displaystyle x = y^{1/a}$