Find $1/x+1/y+1/z$

Question

The values are as follows: $p^{x-1}=qr,r^{z-1}=pq,q^{y-1}=rp$

I have to find: $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$

What I have tried so far:
$$\frac{p^x}{p}=qr,\frac{q^y}{q}=rp,\frac{r^z}{r}=pq$$ $$p^x=pqr,q^y=pqr,r^z=pqr$$

I am not sure if this part is correct. However, it does point to an option - $1$ $$p^x=p^1q^1r^1$$ $$x=1$$ $$\text{Similarly},y=1,z=1$$

$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{1}+\frac{1}{1}+\frac{1}{1}=1$$

I don't think this part is correct. How can I simplify this further?

Also, there are 4 options given:
$$2$$ $$1$$ $$3$$ $$0$$

Answer 1

As you correctly showed, we have : $$p^x = q^y = r^z = pqr$$ This implies that : \begin{align} p &= (pqr)^{1/x} \\ q &= (pqr)^{1/y} \\ r &= (pqr)^{1/z} \end{align} Multiplying the $3$ equations, we get : $$pqr = (pqr)^{1/x+1/y+1/z}$$ I am guessing a hypothesis is that $p,q,r$ are positive real number. If this is the case, we must have : $$\frac1x+\frac1y+\frac1z = 1$$

(Note that this means your result $x=y=z=1$ is false)

Answer 2

I assume $p,q,r \in \mathbb{R}^{+}$

We've $$x=\log_{p}qr +1 =\log_{p}(pqr)$$ $$y=\log_{q}rp +1=\log_{q}(pqr)$$ $$z=\log_{r}pq +1=\log_{r}(pqr) $$

Therefore $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{\ln p +\ln q +\ln r}{\ln pqr}=1$$

Answer 3

$$p^x=pqr, p=pqr^\frac{1}{x}$$ $$r^z=pqr, r=pqr^\frac{1}{z}$$ $$q^y=pqr, q=pqr^\frac{1}{y}$$

$$pqr= pqr^{\frac{1}{x}+\frac{1}{z}+\frac{1}{y}}$$

$$\frac{1}{x}+\frac{1}{z}+\frac{1}{y}=1$$