Question

value of
log
128 to the base8​

Answer 1

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