$\displaystyle Log\sqrt{x^3-3x}=\frac{1}{2}$
I got to: $\displaystyle 10^\frac{1}{2}=(x^3-3x)^\frac{1}{2}$ and then got a little confused on how I would distribute the exponent and solve it.
Any and all pointers highly appreciated, thanks!
$\displaystyle Log\sqrt{x^3-3x}=\frac{1}{2}$
I got to: $\displaystyle 10^\frac{1}{2}=(x^3-3x)^\frac{1}{2}$ and then got a little confused on how I would distribute the exponent and solve it.
Any and all pointers highly appreciated, thanks!
Using Log laws:
$\displaystyle \log(\sqrt{x^3-3x}) = \dfrac{1}{2}\log(x^3-3x)$
$\displaystyle \log(x^3-3x) = 1$
$\displaystyle x^3-3x = 10$
Can you continue? Bear in mind $\displaystyle x > 0$
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Using your method you can square both sides to give $\displaystyle 10 = x^3-3x$ which goes to show there's more than one way to skin a cat. Since we know that $\displaystyle x>0$ we needn't worry about extraneous solutions
Certainly
You have the equation $\displaystyle \log(\sqrt{x^3-3x}) = \dfrac{1}{2}$. By the laws of logarithms $\displaystyle [\ln(a^k) = k\ln(a)]$ so it can be as $\displaystyle \dfrac{1}{2}\log(x^3-3x) = \dfrac{1}{2}$.
Either by inspection or multiplying by 2 will cancel the 1/2 on both sides to give $\displaystyle \log(x^3-3x) = 1$