If log25 ~ 2.3219,
evaluate log280.

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 .

Step-by-step explanation:

d.192

step-by-step explanation:

12*16=192

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