Question

If ${a^x} = b$ , ${b^y} = c$ and ${c^z} = a$ , prove that $xyz = 1$

Answer 1

Hint:
We can take the logarithm of the three given equations. Then we can rearrange them and write them as equations of x, y and z. Then we can take the LHS of the equation that we need to prove and then by substitution, we can prove that LHS is equal to the RHS.

Complete step by step solution:
We have the equations ${a^x} = b$ , ${b^y} = c$ and ${c^z} = a$
Consider ${a^x} = b$ ,
We can log on both sides. Thus, we will get,
 $ \Rightarrow \log {a^x} = \log b$
We know that $\log {a^b} = b\log a$ . Using this relation, we get,
 $ \Rightarrow x\log a = \log b$
Now we can divide the equation throughout with $\log a$ .
  $ \Rightarrow x = \dfrac{{\log b}}{{\log a}}$ …. (1)
Consider ${b^y} = c$ ,
We can log on both sides. Thus, we will get,
 $ \Rightarrow \log {b^y} = \log c$
We know that $\log {a^b} = b\log a$ . Using this relation, we get,
 $ \Rightarrow y\log b = \log c$
Now we can divide the equation throughout with $\log b$ .
  $ \Rightarrow y = \dfrac{{\log c}}{{\log b}}$ …. (2)
Now we can consider ${c^z} = a$ ,
We can log on both sides. Thus, we will get,
 $ \Rightarrow \log {c^z} = \log a$
We know that $\log {a^b} = b\log a$ . Using this relation, we get,
 $ \Rightarrow z\log c = \log a$
Now we can divide the equation throughout with $\log c$ .
  $ \Rightarrow z = \dfrac{{\log a}}{{\log c}}$ …. (3)
We need to prove that $xyz = 1$ .
We can take the LHS.
 $ \Rightarrow LHS = xyz$
We can substitute equations (1), (2) and (3). So, the LHS will become,
 $ \Rightarrow LHS = \dfrac{{\log b}}{{\log a}} \times \dfrac{{\log c}}{{\log b}} \times \dfrac{{\log a}}{{\log c}}$
Now we can cancel the common terms in the numerator and denominator.
 $ \Rightarrow LHS = 1$
But the $RHS = 1$
So, we can write,
 $ \Rightarrow LHS = RHS$
Hence proved.

Note:
Alternate method of solving this problem is given by,
We have the equations,
 ${a^x} = b$ … (a)
 ${b^y} = c$ … (b)
 ${c^z} = a$ …. (c)
On substituting equation (a) in (b), we get.
 $ \Rightarrow {\left( {{a^x}} \right)^y} = c$
We know that ${\left( {{a^x}} \right)^y} = {a^{xy}}$
 \[ \Rightarrow {a^x}^y = c\] … (d)
Now we can substitute equation (d) in (c).
 $ \Rightarrow {\left( {{a^{xy}}} \right)^z} = a$
We know that ${\left( {{a^x}} \right)^y} = {a^{xy}}$
 \[ \Rightarrow {a^x}^{yz} = a\]
As the base of the LHS and RHS are equal, we can equate the powers. So, we get,
 $ \Rightarrow xyz = 1$
This is the equation we need to prove.