How should I go about solving this? Is it even possible? 32/48 = (6^b)/(14^b)
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Originally Posted by mcsquared How should I go about solving this? Is it even possible? 32/48 = (6^b)/(14^b) I hope you know logarithms and the rules of logs so so so yeah... gotta use logarithms
32/48=6^b/14^b 32*14^b=14^b*48 32*2^b*7^b=2^b*3^b*48 32*7^b=3^b*48 32/48=(3/7)^b lg(32/48)=lg(3/7)^b =blg(3/7) b=lg(32/48)/lg(3/7) = 0.479(3s.f)
Keep it SIMPLE: (6/14)^b = 32/48 (3/7)^b = 2/3 b = log(2/3) / log(3/7)
Oh man, that was kinda obvious. I thought to simplify the 6/14 but the exponent being a variable made me doubt myself. D'oh! Thank you much for the solutions, everyone.
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