# 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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The value of `log_2 5` is approximately 2.3219.

`log_2 80`

= `log_2 (5*16)`

= `log_2 5 + log_2 16`

= `log_2 5 + log_2 2^4`

= `log_2 5 + 4*log_2 2`

= `log_2 5 + 4`

= 4 + 2.3219

= 6.3219

**The value of `log_2 80` is 6.3219**