log base b of (5b^2)=?
log base b of ^3√2b=?
Approximate the logarithm, using log base b of 2 is = to about 1.098, log base b of 3 is about 1.740, and log base b 5 is about 2.5495?
2016-05-20 9:08 pm
回答 (3)
2016-05-20 9:37 pm
2016-05-20 9:58 pm
log_b(5b²) = log_b(5) + 2 log_b(b) = log_b(5) + 2 ≈ 4.5495
log_b(³√(2b) = 1/3 log_b(2b)) = 1/3 log_b(2) + 1/3 log_b(b) = 1/3(1.098) + 1/3 ≈ 0.6987
log_b(³√(2b) = 1/3 log_b(2b)) = 1/3 log_b(2) + 1/3 log_b(b) = 1/3(1.098) + 1/3 ≈ 0.6987
2016-05-20 9:16 pm
Log_b (5b^2) = 2 + 2.5495 = 4.5495.
Log_b (b*cube root of 2) = 1 + 0.366 = 1.366.
If you meant Log_b[cube root of (2b)], you should have used parentheses to indicate that; the answer would be (1/3)*(2.098) = 0.699.
Log_b (b*cube root of 2) = 1 + 0.366 = 1.366.
If you meant Log_b[cube root of (2b)], you should have used parentheses to indicate that; the answer would be (1/3)*(2.098) = 0.699.
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