# If log(base 2)5 is approximately 2.3219, evaluate log(base 2)80.Give your answer correct to four decimals.

You need to write `log_2 80` in terms of `log_2 5` , such that:

`log_2 80 = log_2 (8*10) `

Using the properties of logarithms, you need to convert the logarithm of product in summation of logarithms, such that:

`log_2 (8*10)= log_2 8 + log_2 10`

Since `log_2 10 = log_2 (2*5)` , yields:

`log_2 (8*10)= log_2 8 + log_2 (2*5)`

`log_2 (8*10)= log_2 8 + log_2 2 + log_2 5`

You need to write 8 as power of 2, such that:

`log_2 (8*10) = log_2 (2^3) + log_2 2 + log_2 5`

Using the power property of logarithms, `log_a b^c = c*log_a b ` yields:

`log_2 (8*10) = 3log_2 2 + log_2 2 + log_2 5`

Since `log_2 2 = 1` yields:

`log_2 (8*10) = 3 + 1 + 2.3219`

`log_2 (8*10) = 4 + 2.3219 => log_2 (8*10) = 6.3219`

Hence, evaluating the given logarithm , using the indicated properties of logarithms, yields `log_2 (8*10) = 6.3219` .

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