Logarthmic complex!

Geometry Level 4

$\LARGE {\log_{\tan(30^\circ)} \frac{2|z|^{2}+2|z|-3}{|z|+1}<-2}$

then choose the correct option

note that $abs(z)=|z|$

abs(z)>\frac{3}{2}

abs(z)>2

abs(z)<\frac{3}{2}

abs(z)<2

1 solution

Vishwak Srinivasan
Aug 6, 2015

$\tan 30^{\circ} = \dfrac{1}{\sqrt{3}} < 1$

Hence when we remove the logarithm to convert the inequation into a easy solvable inequation, the inequality is reversed.

$\dfrac{2|z^2| + 2|z| - 3}{|z|+1} > \left(\dfrac{1}{\sqrt{3}}\right)^{-2}$

$\dfrac{2|z^2| + 2|z| - 3}{|z| + 1} > 3$

$\Rightarrow 2|z|^2 + 2|z| - 3 > 3|z| + 3 \Rightarrow 2|z|^2 - |z| - 6 > 0$

$(2|z| + 3)(|z| - 2) > 0 \Rightarrow \large\boxed{|z| > 2}$

hey @Vishwak Srinivasan do the inequality always change when the log is moved to the R.H.S??

Palash Som - 5 years, 2 months ago

No,The inequality only changes when the base of logarithm is less than 1,otherwise it would remain the same.

Ayush Agarwal - 5 years, 2 months ago