college algebra - solve
2006-12-06 06:36:24 · 6 answers · asked by j t 1 in Science & Mathematics ➔ Mathematics
Subtract to get log_b (A/B)=log_b (A)-log_b (B)=1.387-.146=1.241
2006-12-06 06:38:58 · answer #1 · answered by mathematician 7 · 1⤊ 0⤋
you must know log (A/B) = log A - log B
so 1.387 -0.146 = 1.241
and A/B = 10^1.241 = 17.4
2006-12-06 14:41:33 · answer #2 · answered by maussy 7 · 0⤊ 0⤋
log b A = 1.387 and log b B = 0.146.
log b (A/B) = log b A - log b B
log b (A/B) = 1.387 - 0.146
log b (A/B) = 1.241
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2006-12-06 14:49:05 · answer #3 · answered by Luiz S 7 · 0⤊ 0⤋
simply use the formula: log(A/B) = log(A) - log (B)
= 1.387 - 0.146 = 1.241
2006-12-06 14:59:35 · answer #4 · answered by red12saleen 2 · 0⤊ 0⤋
log_b A/B = log_b A - log_b B = 1.387 - 0.146 = 1.241
2006-12-06 14:43:54 · answer #5 · answered by steiner1745 7 · 0⤊ 0⤋
log(A/B)=logA-logB
=1.387-0.146
=1.241
2006-12-06 14:38:39 · answer #6 · answered by raj 7 · 0⤊ 1⤋