NDA 2018
Exam NDA
Year 2018
Subject Mathematics
Chapter Basics of Mathematics NDA 2018 · Mathematics · Basics of Mathematics
What is the value of _7 _7 7 7 7 equal to?
What is the value of
log 7 log 7 7 7 7 \log _7 \log _7 \sqrt{7 \sqrt{7 \sqrt{7}}} log 7 log 7 7 7 7 equal to?
Options A 3 log 2 7 3 \log _2 7 3 log 2 7 B 1 − 3 log 2 7 1-3 \log _2 7 1 − 3 log 2 7 C 1 − 3 log 7 2 1-3 \log _7 2 1 − 3 log 7 2 D 7 8 \frac{7}{8} 8 7 Correct answer C. 1 − 3 log 7 2 1-3 \log _7 2 1 − 3 log 7 2
Step-by-step solution
Let
log 7 log 7 7 7 7 = x \log _7 \log _7 \sqrt{7 \sqrt{7 \sqrt{7}}}=x log 7 log 7 7 7 7 = x
then
7 x = log 7 7 7 7 7 x = 1 2 ⋅ log 7 7 7 7 7 x = 1 2 [ log 7 7 + 1 2 log 7 7 7 ] = 1 2 [ 1 + 1 2 log 7 7 + 1 2 log 7 7 ] = 1 2 [ 1 + 1 2 + 1 4 ] 7 x = 7 8 x = log 7 ( 7 8 ) x = log 7 7 − log 7 8 x = 1 − 3 log 7 2 \begin{aligned}
7^x & =\log _7 \sqrt{7 \sqrt{7 \sqrt{7}}} \\
7^x & =\frac{1}{2} \cdot \log _7 7 \sqrt{7 \sqrt{7}} \\
7^x & =\frac{1}{2}\left[\log _7 7+\frac{1}{2} \log _7 7 \sqrt{7}\right] \\
& =\frac{1}{2}\left[1+\frac{1}{2} \log _7 7+\frac{1}{2} \log _7 \sqrt{7}\right] \\
& =\frac{1}{2}\left[1+\frac{1}{2}+\frac{1}{4}\right] \\
7^x & =\frac{7}{8} \\
x & =\log _7\left(\frac{7}{8}\right) \\
x & =\log _7 7-\log _7 8 \\
x & =1-3 \log _7 2
\end{aligned} 7 x 7 x 7 x 7 x x x x = log 7 7 7 7 = 2 1 ⋅ log 7 7 7 7 = 2 1 [ log 7 7 + 2 1 log 7 7 7 ] = 2 1 [ 1 + 2 1 log 7 7 + 2 1 log 7 7 ] = 2 1 [ 1 + 2 1 + 4 1 ] = 8 7 = log 7 ( 8 7 ) = log 7 7 − log 7 8 = 1 − 3 log 7 2
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