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OpenStudy (anonymous):
how can i evaluate log2(thats a little 2)160-log2(little 2)5 without using a calculator?
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OpenStudy (anonymous):
ok, so when you subtract to logs, it's like dividing the numbers you are taking the logs of
OpenStudy (anonymous):
so, log2(x) - log2(y) is the same as log2(x/y)
OpenStudy (anonymous):
log(base 2)160-log(base 2)5=log(base 2)32, as mindedone said. therefore, ask yourself 2 to what power is 32.
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