GATE CSE 2018 | Question: GA-7

Question

 If $pqr \ne 0$ and $p^{-x}=\dfrac{1}{q},q^{-y}=\dfrac{1}{r},r^{-z}=\dfrac{1}{p},$ what is the value of the product $xyz$ ?

• A. $-1$
• B. $\dfrac{1}{pqr}$
• C. $1$
• D. $pqr$

Comments

got 1

---

same here ans should be 1

---

you have type wrong question

p−x=1/q

q−y=1/r

r−z=1/p

---

pqr !=0, p^-x = 1/q, q^-y = 1/r, r^-z = 1/p

Taking log both sides,

-x(log p) = -log q,

-y(log q) = -log r,

-z(log r) = -log p

=> x = log q/ log p

=> y = log r/ log q

=> z = log p/ log r

multiplying x,y and z,

(x)(y)(z) = 1

Answer 1

$\color{maroon}{pqr \neq 0}$

$\color{maroon}{p^{-x} = \dfrac{1}{q}}$

Or,$\log(p^{-x}) = \log(\dfrac{1}{q})$

Or, $-x \log (p) = \log(1)-\log(q)$  $\qquad\Big[∵\color{blue}{\log_b(m^n)= n. \log_b(m)\\\qquad\qquad \log_b(\dfrac{m}{n})= \log_b(m)-\log_b(n)} \Big]$

Or, $x\log(p) = \log (q)$ $\qquad[∵\color{blue}{\log(1)=0}]$

Or, $\color{green}{x= \dfrac{\log(q)}{\log(p)}}$

$\color{maroon}{q^{-y} = \dfrac{1}{r}}$

Or,$\log(q^{-y}) = \log(\dfrac{1}{r})$

Or, $-y \log (q) = \log(1)-\log(r)$  $\qquad\Big[∵\color{blue}{\log_b(m^n)= n. \log_b(m)\\\qquad\qquad \log_b(\dfrac{m}{n})= \log_b(m)-\log_b(n)}\Big]$

Or, $y\log(q) = \log (r)$ $\qquad[∵\color{blue}{\log(1)=0}]$

Or, $\color{green}{y= \dfrac{\log(r)}{\log(q)}}$

$\color{maroon}{r^{-z} = \dfrac{1}{p}}$

Or,$\log(r^{-z}) = \log(\dfrac{1}{p})$

Or, $-z \log (r) = \log(1)-\log(p)$  $\qquad\Big[∵\color{blue}{\log_b(m^n)= n. \log_b(m)\\\qquad\qquad \log_b(\dfrac{m}{n})= \log_b(m)-\log_b(n)}\Big]$

Or, $z\log(r) = \log (p)$ $\qquad[∵\color{blue}{\log(1)=0}]$

Or, $\color{green}{z= \dfrac{\log(p)}{\log(r)}}$

∴ $\color{black}{x \times y \times z}$ = $ \dfrac{\log(q)}{\log(p)} \times  \dfrac{\log(r)}{\log(q)} \times \dfrac{\log(p)}{\log(r)}$

$\qquad \qquad = \color{black}{1}$

Correct Answer: $C$

Comments

$p^{-x}=\dfrac{1}{q},q^{-y}=\dfrac{1}{r},r^{-z}=\dfrac{1}{p}$

We can write like this $\dfrac{1}{p^{x}}=\dfrac{1}{q},\dfrac{1}{q^{y}}=\dfrac{1}{r},\dfrac{1}{r^{z}}=\dfrac{1}{p}$

$\implies \dfrac{1}{p^{x}}\times \dfrac{1}{q^{y}}\times \dfrac{1}{r^{z}} = \dfrac{1}{q}\times\dfrac{1}{r} \times \dfrac{1}{p}$

$\implies \dfrac{1}{p^{x}}\times \dfrac{1}{q^{y}}\times \dfrac{1}{r^{z}} = \dfrac{1}{p}\times\dfrac{1}{q} \times \dfrac{1}{r}$

Compare the coefficients of both sides, and we get,

$\implies x=y=z=1$

$\therefore xyz = 1\times 1 \times 1 = 1$

So, the correct answer is $(C).$

---

A simple and quick soluion 

p^-x = 1/q   => 1/p^x =1/q   =>  p^x=q   ------(i)

q^-y = 1/r   => 1/q^y =1/r   =>  q^y=r     ------(ii)

r^-z = 1/p   => 1/r^z =1/p   =>  r^z=p     ------(iii)

multiplying i,ii,iii

pqr=p^xq^yr^z

^{taking log both the sides}

log(pqr)= log(p^xq^yr^z)

logp + logq + logr = xlogp + ylogq +zlogr

comparing the coefficients in lhs and rhs 

x=1,y=1,z=1

xyz=1.1.1=1

Answer 2

Given: $pqr ≠ 0$

$p^{-x}= 1/q$

$q^{-y}= 1/r$

$r{-z}= 1/p$

Take $\log$ on both side

$x \log p= \log q$

$\implies x= \frac{\log q}{\log p}$

$y \log q= \log r$

$\implies y = \frac{\log r}{ \log q}$

$z \log r= \log p$

$\implies z= \frac{\log p}{\log r}$

So, $xyz= 1.$

Comments

Best.

Answer 3

Here we can write as follows:

p^x=q --(i)

q^y=r --(ii)

r^z=p --(iii)

from (ii) and (iii)

(q^y)^z=p => q^yz=p --(iv)

from (i) and (iv)

(p^x)^yz=p => p^xyz=p¹   

              => xyz=1

 

 

Comments

Thankyou for the simplest explanation!!!

Answer 4

multiply each equation

 

p^-x *q^-y*r^-z=1/pqr

on comparing

x=-1

y=-1

z=-1

so xyz=1

Comments

yes got same

---

if p=q=r=1 then pqr!=0 it can take any  positive value of x,y,z.