(x)^4/3 = 16

How do I find this?

Oh and more importantly, how do you do this?

(9)^x = 4th root (27) / 3

Printable View

- Jul 23rd 2007, 09:51 AMJonathanEyoonExponent Question...
(x)^4/3 = 16

How do I find this?

Oh and more importantly, how do you do this?

(9)^x = 4th root (27) / 3 - Jul 23rd 2007, 10:04 AMtopsquark
$\displaystyle x^{4/3} = 16$ <-- Cube both sides

$\displaystyle x^4 = 16^3$ <-- Take the fourth root of both sides

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

What number is this? Well if you recall that $\displaystyle 2^4 = 16$...

$\displaystyle x = 16^{3/4} = \left ( 2^4 \right ) ^{3/4} = 2^{4 \cdot (3/4)} = 2^3 = 8$

Of course, this is the principle value of the solution. You might note that x = -8 also solves the equation. (The easiest way to see this is to take $\displaystyle x^4 = 16^3$ and take the square root of both sides twice.) There are also two complex solutions. I can give you the general solution if you like; it's not much harder than what I derived here.

-Dan - Jul 23rd 2007, 10:09 AMtopsquark
You want to get both sides of the equation to the same base. In this case, the easiest base to use is 3.

$\displaystyle 9 = 3^2$ and $\displaystyle 27 = 3^3$, and $\displaystyle \sqrt[4]{x} = x^{1/4}$, so...

$\displaystyle 9^x = \frac{\sqrt[4]{27}}{3}$

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

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

Now we have that

$\displaystyle 2x = -\frac{1}{4}$

$\displaystyle x = -\frac{1}{8}$

-Dan