How do you solve #log_8 48 - log_8 w = log_8 4#?

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Answer 1 · 2016-08-09T12:49:17

#w = 12#

We can write the left hand side as a single log term.

#If the logs are being subtracted, the numbers are being divided"

#log_8 48 - log_8 w = log_8 4#

#log_8 (48/w) = log_8 4#

#:. 48/w = 4" if log A = log B, then " A = B#

#48 = 4w#

#w = 12#

(Luckily all the logs had the same base.)