GATE Mechanical 2018 Set 2 | GA | Question: 5

Question

The value of the expression $\dfrac{1}{1+ \log_u \: vw} + \dfrac{1}{1+ \log_v \: wu} + \dfrac{1}{1+\log_w uv}$ is _______

• A. $-1$
• B. $0$
• C. $1$
• D. $3$

Comments

👉 Quick trick: whenever such cyclic log expressions appear symmetrically, you can safely assume powers of a single base (like $2, 4, 8$ ).

 

 

Take \( u = 2, \, v = 4, \, w = 8 \). These are powers of 2, so logs become neat.
\[ \log_u(vw) = \log_2(4 \times 8) = \log_2(32) = 5 \]
Hence the first term = \( \frac{1}{1 + 5} = \frac{1}{6} \)
\[ \log_v(wu) = \log_4(8 \times 2) = \log_4(16) = 2 \]
Hence the second term = \( \frac{1}{1 + 2} = \frac{1}{3} \)
\[ \log_w(uv) = \log_8(2 \times 4) = \log_8(8) = 1 \]
Hence the third term = \( \frac{1}{1 + 1} = \frac{1}{2} \)
Now add them all:
\[ \frac{1}{6} + \frac{1}{3} + \frac{1}{2} = \frac{1 + 2 + 3}{6} = 1 \]
Final Answer = 1
So the correct option is (C) 1.

Answer 1

$\dfrac{1}{1+ \log_{u} \: vw} + \dfrac{1}{1+ \log_{v} \: wu} + \dfrac{1}{1+\log_{w} uv}$

$ = \dfrac{1}{\log_{u} \: u + \log_{u} \: vw} + \dfrac{1}{\log_{v} \: v  + \log_{v} \: wu} + \dfrac{1}{\log_{w} \: w +\log_{w} \: uv}$

$ = \dfrac{1}{\log_{u} \: uvw} + \dfrac{1}{\log_{v} \: vwu} + \dfrac{1}{\log_{w} \: wuv}$

$ = \dfrac{1}{\log_{u} \: uvw} + \dfrac{1}{\log_{v} \: uvw} + \dfrac{1}{\log_{w}\: uvw}$

$ = \log_{uvw} \: u + \log_{uvw} \: v + \log_{uvw} \: w$

$ = \log_{uvw} \: uvw $

$ = 1\qquad\because\left(\log_{a} \: a = 1 \right)$

Hence $(C)$ is Correct.

Comments

assume  u=v=w=10

we will get ans as 1

---

why just u=v=w=10 only…

you can also take

u=v=w=100

u=v=w=1000

u=v=w=2

u=v=w=3

.

.

u=v=w =K

K is any number for which log is defined.

 

---

@suchithreddy and @Kiyoshi  by assuming $(u=v=w)$, you are decreasing the domain of this problem. this approach can only work in certain type of problems and if you are lucky then only you can think of such a brute-force-ish method and get a solution in a resonable amount of time during exam. 

---

nice approach.