Question

If x=log9/7, y = log'7/5, z =log5/3. Show that xyz = 2​

Answer 1

Answer:

Answer:

$\sf \: xyz=3$

Step-by-step explanation:

Formula used:

Base changing rule:

$\sf \: log_a{c}=(log_a{b})(log_b{c})$

Power rule:

$\sf \: log_aM^n=n\:log_am$

$\sf \: log_a{a}=1$

Given:

$\sf \: x=log_5{7}$

$\sf \: y=log_7{27}$

$\sf \: z=log_3{5}$

Now,

$\sf \: xyz=(log_5{7})(log_7{27})(log_3{5}) \:$

$\sf \: xyz=(log_5{27})(log_3{5})$

$\sf \: xyz=(log_5{3^3})(log_3{5})$

$\sf \: xyz=3(log_5{3}) \: \: (log_3{5})xyz=3$

$\sf \: xyz=3(log_5 {5})$

$\sf \: xyz=3 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...(1)$

xyz=3

Hence proved