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If log(base 2)5 is approximately 2.3219, evaluate log(base 2)80. Give your answer...

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kristenmarieb... | Student, Grade 10 | Valedictorian

Posted August 26, 2013 at 2:04 PM via web

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If log(base 2)5 is approximately 2.3219, evaluate log(base 2)80.

Give your answer correct to four decimals.

Tagged with algebra2, math

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llltkl | College Teacher | Valedictorian

Posted August 26, 2013 at 2:25 PM (Answer #1)

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Given `log_2(5)=2.3219`

To evaluate `log_2(80)`

`=log_2(5*16)`

As per the product rule of logarithms, `log_a(m*n)=log_am+log_an` . So, we can write:

`=log_2(5)+log_2(16)`

`=log_2(5)+log_2(2^4)`

Now, use the exponent rule of logarithms i.e `log_am^n=nlog_am` .

`=log_2(5)+4log_2(2)`

`=2.3219+4`       (since `log_2(2)=1` )

`=6.3219`

Therefore, `log_2(80)=6.3219.`

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