Asked by help please
The value of log16(0.0036) can be rounded to two decimal places. What is the approximate value? (Hint: log10(2)=0.301, log10(3)=0.477)
Answers
Answered by
oobleck
log_16(x) = logx/log16
so log_16(0.0036) = log(0.0036)/log16 = (2(log2+log3)-4)/(4log2)
now plug in your values
so log_16(0.0036) = log(0.0036)/log16 = (2(log2+log3)-4)/(4log2)
now plug in your values
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