value of
log
128 to the base8
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Answer:
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Step-by-step explanation:
Given,
\log_8 128log
8
128
We have to find, the value of \log_8 128log
8
128 is:
Solution:
∴ \log_8 128log
8
128
= \log_8 (64\times 2)log
8
(64×2)
Using the logarithm identity:
\log (a\times b)log(a×b) = \log aloga + \log blogb
= \log_8 64+\log_8 2log
8
64+log
8
2
= \log_8 8^2+\log_{2^3} 2log
8
8
2
+log
2
3
2
Using the logarithm identity:
\log a^bloga
b
= b\log aloga and
\log_{a^c} alog
a
c
a = \dfrac{1}{c} \log_aa
c
1
log
a
a
= 2\log_8 8+\dfrac{1}{3} \log_{2} 22log
8
8+
3
1
log
2
2
= 2 + \dfrac{1}{3}
3
1
[ ∵ \log_{a} alog
a
a = 1]
= \dfrac{6+1}{3}
3
6+1
= \dfrac{7}{3}
3
7
∴ \log_8 128log
8
128 = \dfrac{7}{3}
3
7
Thus, the required "option a) \dfrac{7}{3}
3
7
" is correct
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