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

Give your answer correct to four decimals.

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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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