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Question : 60 of 160
Marks:
+1,
-0
Solution:
‌log42−log82+log162−...‌‌=‌−‌+‌−...‌{∵logba=‌}‌‌=‌−‌+‌−...‌‌=‌−‌+‌−...‌‌‌=(‌−‌+‌−...)‌∵loge(1+x)=x−‌+‌−‌+....1) (Put
x=1 )
‌⇒loge(1+1)=1−‌+‌−‌+....‌⇒‌−‌+‌−...=1−loge2‌‌‌=1−loge2
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