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Evaluate log_2 1/8 + log_5 125 + log_27 9show complete manual solution

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spock1 | Student, Undergraduate | Honors

Posted May 26, 2012 at 7:23 PM via web

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Evaluate log_2 1/8 + log_5 125 + log_27 9

show complete manual solution

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thilina-g | College Teacher | (Level 1) Educator

Posted May 26, 2012 at 7:30 PM (Answer #1)

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`log_2 1/8 + log_5 125 + log_27 9`

We can rewrite,

`1/8 = 1/2^3 = 2^(-3)` and

`125 = 5^3`

`9 = 3^2`   But we know,

`27 = 3^3`

Therefore,

`3 = 27^(1/3)`

Therefore, `9 = (27^(1/3))^2 = 27^(2/3)`

Therefore,

`log_2 1/8 + log_5 125 + log_27 9 = log_2 2^(-3) + log_5 5^2 + log_27 27^(2/3)`

This gives,

`log_2 1/8 + log_5 125 + log_27 9 = -3+2+2/3`

`log_2 1/8 + log_5 125 + log_27 9 = -1/3`

 

Therefore the answer is `-1/3` .

 

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