wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Solve loga+logbcloga(1+logabc)

Open in App
Solution

loga+logbcloga(1+logabc)
Changing base of logabc
logabc=logbcloga
Thus, loga+logbcloga(1+logabc)=loga+logbcloga(1+logbcloga)
loga+logbcloga(loga+logbcloga)
loga+logbc(loga+logbc)
1

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Integration by Substitution
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon