Hello...I have completely forgotten how do solve for x in questions like this

(sorry I don't know how to put 2/3 as a power)

x to the 2/3 =9

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- Nov 10th 2007, 09:42 PMdalshaexponential equations
Hello...I have completely forgotten how do solve for x in questions like this

(sorry I don't know how to put 2/3 as a power)

x to the 2/3 =9 - Nov 10th 2007, 09:54 PMearboth
Hello,

rearrange your equation:

$\displaystyle x^{\frac23} = 9$ is the same as: $\displaystyle \sqrt[3]{x^2} = 9

$

1. Get rid of the root: $\displaystyle \left(\sqrt[3]{x^2}\right)^3 = 9^3~\iff~x^2 = 9^3$

2. Since $\displaystyle 9 = 3^2$ you have now: $\displaystyle x^2 = 3^6~\iff~x=3^3 = 27~\vee~x=-27$ - Nov 10th 2007, 10:22 PMangel.white
**Method1:**

1. Initial Equation

$\displaystyle x^{\frac{2}{3}}=9$

2. Take each side to the $\displaystyle \frac{3}{2}$ power

$\displaystyle (x^{\frac{2}{3}})^{\frac{3}{2}}=9^{\frac{3}{2}}$

3. Because $\displaystyle (a^{b})^{c}=a^{bc}$

$\displaystyle x^{\frac{2}{3}\cdot \frac{3}{2}}=9^{\frac{3}{2}}$

4. Simplify

$\displaystyle x^{1}=9^{\frac{3}{2}}$

5. Any number to the power of 1 is itself

$\displaystyle x=9^{\frac{3}{2}}$

6. A fractional exponent means you take the number to the power in the numerator, and to the root in the denominator, so $\displaystyle 9^{\frac{3}{2}}=\sqrt{9^3}$ So lets cube 9

$\displaystyle x=729^{\frac{1}{2}}$

7. And square root it

$\displaystyle x=27^{1}$

8. And simplify

$\displaystyle x=27$

-----

**Method2:**1. Initial Equation

$\displaystyle x^{\frac{2}{3}}=9$

2. Take both sides to the natural log (any log will work, but natural log is most common, with the log base ten as the second most common)

$\displaystyle ln(x^{\frac{2}{3}})=ln(9)$

3. One of the properties of logarithms with exponents is that $\displaystyle ln(a^{b}) = b\cdot \ln{a}$, so:

$\displaystyle \frac{2}{3}\ln(x)=ln(9)$

4. Now multiply both sides by 3

$\displaystyle 2\ln(x)=3\ln(9)$

5. And divide both sides by 2

$\displaystyle ln(x)=\frac{3}{2}\ln(9)$

7. And by the same property of logs and exponents we used in step 3

$\displaystyle ln(x)=\ln(9^{\frac{3}{2}})$

8. Now, if a=b then: Any number is equal to itself so$\displaystyle e^{a}=e^{a}$ And because a=b, we can substitute b in for a $\displaystyle e^{a}=e^{b}$ So we can say e to the left hand side, equals e to the right hand side

$\displaystyle e^{ln(x)}=e^{\ln(9^{\frac{3}{2}})}$

9. And because ln is the log base e, lets set ln(x) equal to y, so $\displaystyle ln(x)=y \mbox{ means } e^{y}=x$ And because we are taking $\displaystyle e^{ln(x)}$, we can substitute y in for ln(x): $\displaystyle e^{ln(x)}=e^{y}$. And as we showed earlier, $\displaystyle e^{y}=x$, so, all that means that $\displaystyle e^{ln(x)} = x$ This is true for any x, so lets simplify our equation

$\displaystyle x=9^{\frac{3}{2}}$

10. And now simplify the right hand side using the same method we showed in steps 6-8 from Method 1

$\displaystyle x=27$

Note: you could have done this with any base log you wanted, if the "e" confuses you, try using regular log base ten. - Nov 10th 2007, 10:40 PMJhevon
excellent job, angel.white, just a small note here. since, $\displaystyle a^{bc} =a^{cb}$, we can interpret $\displaystyle 9^{\frac 32}$ as $\displaystyle \left( 9^3 \right)^{\frac 12}$ or $\displaystyle \left( 9^{\frac 12} \right)^3$. you decided to do the first, but it would have been easier to do the second

$\displaystyle 9^{\frac 32} = \left( 9^{\frac 12} \right)^3 = 3^3 = 27$

(finding the cube of 3 is a lot easier than finding the square root of 729 if you don't have the aid of a calculator)

also, here. since the logarithm function is one to one, we can just drop the logs. $\displaystyle \ln x = \ln 9^{3/2} \implies x = 9^{3/2}$

recall that one to one functions have the property: $\displaystyle f(x) = f(y) \implies x = y$ - Nov 10th 2007, 11:22 PMangel.white
Right, it just seemed easier to do the exponent before the root than to explain why order didn't matter :P

I seem to recall an excellent explanation of how fractional exponents work

Oh here it is http://www.mathhelpforum.com/math-help/51855-post3.html ^_^

Wow, I hadn't thought of that, that's much easier than my explanation (though I must admit, I did enjoy figuring out how to explain it, I just wish I could made it as clear in my post as it is in my head)

I am curious why it matters that the equations be one to one. - Nov 10th 2007, 11:38 PMJhevon
i guess. you could just say multiplication is commutative, but maybe that would be complicated

Quote:

I seem to recall an excellent explanation of how fractional exponents work

Oh here it is http://www.mathhelpforum.com/math-help/51855-post3.html ^_^

Quote:

Wow, I hadn't thought of that, that's much easier than my explanation (though I must admit, I did enjoy figuring out how to explain it, I just wish I could made it as clear in my post as it is in my head)

Quote:

I am curious why it matters that the equations be one to one.

- Nov 11th 2007, 01:28 AMangel.white
- Nov 11th 2007, 01:36 AMJhevon
yes, but in doing so you often introduce extraneous solutions when both sides are not the same sign (whenever you use the technique of squaring both sides, you must chack your solutions in the original equation, because chances are, one of them won't work).

note that we can have x^2 = y^2 without x being equal to y. consider x = 1 and y = -1