Question

If ${\log }_{2a}a=x$, ${\log }_{3a}2a=y$ and ${\log }_{4a}3a=z$, then $xyz-2yz$ is equal to

• A. 1
• B. -1
• C. 0
• D. 2

Answer 1

Answer: (C)

0

We have,

${\log }_{2a}a=x,{\log }_{3a}2a=y,{\log }_{4a}3a=z$

${\log }_{2a}a=x$

$\frac{\log a}{\log 2a}=x$

Similarly,

${\log }_{3a}2a=y$

$\frac{\log 2a}{\log 3a}=y$

Similarly,

${\log }_{4a}3a=z$

$\frac{\log 3a}{\log 4a}=z$

Therefore,,

$=xyz-2yz$

$=\frac{\log a}{\log 2a}\times \frac{\log 2a}{\log 3a}\times \frac{\log 3a}{\log 4a}-2\times \frac{\log 2a}{\log 3a}\times \frac{\log 3a}{\log 4a}$

$=\frac{\log a}{\log 4a}-2\times \frac{\log 2a}{\log 4a}$

$=\log a-\log 4-\log a-2\times (\log 2a-\log 4a)$

$=-\log 4-2\times (\log 2a-\log 2a-\log 2)$

$=-\log 4+2\log 2$

$=-\log 4+\log 4$

$=0$

So, the value is 0.