Question

maths TB. ch2 pg2.8 Q39.
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Answer 1

Solution ..........

Answer 2

ANSWER:-

$\bf \: c)2yz$

$STEP \: BY \: STEP \: VERIFICATION:-$

GIVEN,

$x = log_{24}(12) \\ y = log_{36}(24) \\ z = log_{48}(36)$

NOW,

$xyz + 1 \\ = zyx + 1 \\ = log_{48}(36) . log_{36}(24) . log_{24}(12) + 1 \\ = log_{48}(12) + 1 \: \: \: \: \: |since. log_{ \alpha }( \beta ). log_{ \beta }( \gamma ) = log_{ \alpha }( \gamma ) | \\ = log_{48}(12) + log_{48}(48) \\ = log_{48}(576) \\ = { log_{48}(24) }^{2} \\ = 2 log_{48}(24) \\ = 2 log_{48}(36) . log_{36}(24) \\ = 2zy \\ = 2yz$

So, C is correct answer.