# Math Help - How would I simplify this?

1. ## How would I simplify this?

$Log\sqrt{x^3-3x}=\frac{1}{2}$

I got to: $10^\frac{1}{2}=(x^3-3x)^\frac{1}{2}$ and then got a little confused on how I would distribute the exponent and solve it.

Any and all pointers highly appreciated, thanks!

2. Using Log laws:

$\log(\sqrt{x^3-3x}) = \dfrac{1}{2}\log(x^3-3x)$

$\log(x^3-3x) = 1$

$x^3-3x = 10$

Can you continue? Bear in mind $x > 0$

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Using your method you can square both sides to give $10 = x^3-3x$ which goes to show there's more than one way to skin a cat. Since we know that $x>0$ we needn't worry about extraneous solutions

3. Originally Posted by e^(i*pi)
Using Log laws:

$\log(\sqrt{x^3-3x}) = \dfrac{1}{2}\log(x^3-3x)$

$\log(x^3-3x) = 1$
Hmm, do you mind showing me how you got to $\log(x^3-3x) = 1$?

Originally Posted by e^(i*pi)
Using your method you can square both sides to give $10 = x^3-3x$ which goes to show there's more than one way to skin a cat. Since we know that $x>0$ we needn't worry about extraneous solutions
Ah, that would work. Thanks a lot!

4. Originally Posted by youngb11
Hmm, do you mind showing me how you got to $\log(x^3-3x) = 1$?

Ah, that would work. Thanks a lot!
Certainly

You have the equation $\log(\sqrt{x^3-3x}) = \dfrac{1}{2}$. By the laws of logarithms $[\ln(a^k) = k\ln(a)]$ so it can be as $\dfrac{1}{2}\log(x^3-3x) = \dfrac{1}{2}$.

Either by inspection or multiplying by 2 will cancel the 1/2 on both sides to give $\log(x^3-3x) = 1$

5. Originally Posted by e^(i*pi)
Certainly

You have the equation $\log(\sqrt{x^3-3x}) = \dfrac{1}{2}$. By the laws of logarithms $[\ln(a^k) = k\ln(a)]$ so it can be as $\dfrac{1}{2}\log(x^3-3x) = \dfrac{1}{2}$.

Either by inspection or multiplying by 2 will cancel the 1/2 on both sides to give $\log(x^3-3x) = 1$
Thanks a lot for the clear explanation! Have a great day