In triangle ABC, $m\angle{A}=a\log{x}$ $m\angle{B}=a\log{2x}$ $m\angle{C}=a\log{4x}$. what is $m\angle{B}$

Question

In triangle ABC, $m\angle{A}=a\log{x}$ $m\angle{B}=a\log{2x}$ $m\angle{C}=a\log{4x}$. what is $m\angle{B}$

My steps

$a\log{x}+a\log{2x}+\log{4x}=\pi$

which becomes $a\log{x^a}+\log{2x^a}+\log{4x^a}=\pi$

which devolves into: $\log(2^ax^{3a})$

from then I don't know what to do

btw angle B is supposed to be $\frac{\pi}{3}$

Answer 1

$\begin{array}\\ \pi &=a\log{x}+a\log{2x}+a\log{4x}\\ &=\log{x^a}+\log{(2x)^a}+\log{(4x)^a}\\ &=\log{x^a2^ax^a4^ax^a}\\ &=\log{x^{3a}8^a}\\ &=\log{x^{3a}2^{3a}}\\ &=a\log{x^{3}2^{3}}\\ &=a\log{(2x)^{3}}\\ &=3a\log{2x}\\ &=3m\angle B\\ \end{array} $

That was my initial proof.

This may be simpler.

$\begin{array}\\ \pi &=a\log{x}+a\log{2x}+a\log{4x}\\ &=a(\log{x}+\log{4x})+a\log{2x}\\ &=a\log{4x^2}+a\log{2x}\\ &=2a\log{2x}+a\log{2x}\\ &=3a\log{2x}\\ &=3m \angle B\\ \end{array} $

Nice that they agree.

Answer 2

$a\log{x}+a\log{2x}+a\log{4x}= \pi$

$=a(\log(8x^3))=a(\log(2x)^3)=3a(\log(2x))$

Therefore, angle $B = a(\log(2x))={\pi\over 3}$