Question

log23*log34*log45*log56*log67*log78*l-3
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Answer 1

According to logarithm;

log_ab = (log_kb)/(log_ka)

If we consider k = 10 which is log in base 10 then;

log_ab = (logb)/(loga)

Similarly we can write;

(log2(3))(log3(4))(log4(5))(log5(6))(log6(7))(log7(8))

= (log3)/(log2)xx(log4)/(log3)xx(log5)/(log4)xx........(log7)/(log6)xx(log8)/(log7)

= (log8)/(log2)

= (log2^3)/(log2)

= (3log2)/(log2)

= 3