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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.

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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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tonys538 | Student

The value of `log_2 5` is approximately 2.3219. The value of `log_2 80` has to be determined.

`log_2 80`

= `log_2(16*5)`

= `log_2(2^4*5)`

Use the property log a*b = log a + log

= `log_2 2^4 + log_2 5`

Use the property log a^b = b*log a

= `4*log_2 2 + log_2 5`

Use the property `log_b b = 1`

= 4 + 2.3219

= 6.3219

The required value of `log_2 80 = 6.3219`

Zaca | Student

One rule for Logarithms is that `logx*b = logx + logb`

(I will be writing log, assuming base 2 for all logs)

Using the above rule, 

`log80 = log(5*16) = log5 + log16 = 2.3219 + 4`

Answer: 6.3219