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In the below picture how does he get fromLog 10 Q 100 =3.4893 to Q 100 = 3085m 3 /s When I use e 3.4893

In the below picture how does he get fromLog10Q100=3.4893 to Q100 = 3085m3/s

When I use e3.4893 I get a different answer.

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m =2.2 and s = 0.480 (To be used in part b) log on = 2.2+ K ) 0.480 (6) K (0.5) =2.686 Substitute in equation 6 logi Cian = 3.4893 Cun = 3085m /s

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