CAT 2018 Slot 1 — QA Question 23

log₁₂81 = p

⇒ p = 4log₁₂3

Now, $\frac{4-p}{4+p}=\frac{4-4{\log }_{12}3}{4+4{\log }_{12}3}$

= $\frac{1-{\log }_{12}3}{1+{\log }_{12}3}$

= $\frac{{\log }_{12}12-{\log }_{12}3}{{\log }_{12}12+{\log }_{12}3}=\frac{{\log }_{12}12/3}{{\log }_{12}12\times 3}=\frac{{\log }_{12}4}{{\log }_{12}36}$

= $\frac{2{\log }_{12}2}{2{\log }_{12}6}$ = ${\log }_{6}2$

Hence, $3\left(\frac{4-p}{4+p}\right)=3{\log }_{6}2={\log }_{6}8$

Alternately,
log₁₂81 = p
⇒ 81 = 12^p
⇒ 3⁴ = 3^p × 2^2p 
​​​​​​​⇒ 3^{(4 − p)} = 2^2p

Taking log on both the sides,

(4 – p) (log 3) = (2p) (log 2)

∴ $\frac{\log 3}{\log 2}$ = $\frac{2p}{4-p}$

∴ $\frac{\log 3+\log 2}{\log 2}$ = $\frac{2p + (4-p)}{4-p}$ …(by adding 1 both sides)

∴ $\frac{\log 6}{\log 2}$ = $\frac{4+p}{4-p}$

∴ $\frac{4-p}{4+p}$ = $\frac{\log 2}{\log 6}$ = log₆2

∴  $3\left(\frac{4-p}{4+p}\right)$ = 3log₆2 = log₆8

Hence, option (c).