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Solve:
Log 3/4log 8 (x2+7) +log1/2 log1/4(x2 +7)-1 =-2
5 years ago

Sumit Majumdar
IIT Delhi
137 Points

Dear student,
Please confirm if the question is:
$\left [ log_{3/4}log_{8} \right \left ( x^{2}+7 \right )]+\left [ log_{1/2}log_{1/4} \right \left ( x^{2}+7 \right )^{-1}]=-2$
Then you can simplifiy this as follows:
$\left [ log_{3/4}\left ( \frac{log\left ( x^{2}+7 \right )}{3log2} \right ) \right ]+\left [ log_{1/2}\left ( \frac{log\left ( x^{2}+7 \right )}{2log2} \right ) \right ]=-2$
If we assume:
$u=\frac{log\left ( x^{2}+7 \right )}{log2}$
Then the bove expression can be solved easily.
Regards
Sumit
5 years ago
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### Course Features

• 731 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions