Question

If ${{\log }_{a}}b={{\log }_{b}}c={{\log }_{c}}a$, then prove that $a=b=c$.

Answer 1

Hint: We first try to assume the value of the given equation ${{\log }_{a}}b={{\log }_{b}}c={{\log }_{c}}a$. From the variable we try to find a relation between three unknowns a, b, c and their logarithmic values. We solve the relation involving the value of the equation. We get the value and use that to prove $a=b=c$.

Complete step by step answer:
Let’s assume that ${{\log }_{a}}b={{\log }_{b}}c={{\log }_{c}}a=k$.
We know that ${{\log }_{a}}b=\dfrac{\log b}{\log a}$. Now we multiply ${{\log }_{a}}b$ and ${{\log }_{b}}c$.
So, $\left( {{\log }_{a}}b \right)\times \left( {{\log }_{b}}c \right)=k\times k$. We use the theorem ${{\log }_{a}}b=\dfrac{\log b}{\log a}$.
$\begin{align}
  & \Rightarrow \left( \dfrac{\log b}{\log a} \right)\times \left( \dfrac{\log c}{\log b} \right)={{k}^{2}} \\
 & \Rightarrow \dfrac{\log c}{\log a}={{k}^{2}} \\
\end{align}$
We also have ${{\log }_{c}}a=k$ which means $\dfrac{\log a}{\log c}=k$.
We can see $\dfrac{\log c}{\log a}={{k}^{2}}$ and $\dfrac{\log a}{\log c}=k$ are inverse to each other. This relation gives us that $k=\dfrac{1}{{{k}^{2}}}$.
We solve the equation to get ${{k}^{3}}=1$ which gives us $k=1$.
So, ${{\log }_{a}}b={{\log }_{b}}c={{\log }_{c}}a=1$. We have the identity if ${{\log }_{x}}y=z$ then ${{x}^{z}}=y$.
We apply the theorem and get ${{a}^{1}}=b\Rightarrow a=b$ and ${{b}^{1}}=c\Rightarrow b=c$. This gives $a=b=c$.
Thus proved.

Note: The equation ${{k}^{3}}=1$ is a cubic equation. So, it has 3 solutions which are $k=1,\omega ,{{\omega }^{2}}$. The value of $\omega $ is $\omega =\dfrac{-1\pm \sqrt{3}i}{2}$. The other two values $k=\omega ,{{\omega }^{2}}$ are both complex values. For any value of x, $\log x\in \mathbb{R}$. The other two values get eliminated automatically.