This simplification might be suprisingly easy – if you know the change of base<\/strong> rule<\/p>\n loga<\/sub>b=logb\/loga<\/p>\n as well as the exponent to multiple<\/strong> log rule<\/p>\n loga^b=bloga<\/p>\n and the radical to exponent rule<\/strong><\/p>\n Let’s proceed, starting with the original<\/p>\n log√(27)<\/sub>5<\/sup>√(81)<\/p>\n which becomes, using change of base<\/strong>,<\/p>\n log5<\/sup>√(81)\/log√(27)<\/p>\n Notice that 81=3^4, while 27=3^3. We can rewrite the expression as<\/p>\n log5<\/sup>√3^4}{log√3^3}\\]<\/p>\n Next it becomes, by the radical to exponent rule<\/strong>,<\/p>\n log3^(4\/5)\/log3^(3\/2)<\/p>\n Now, using the exponent to multiple<\/strong> rule, we arrive at<\/p>\n (4log3\/5)\/(3log3\/2)<\/p>\n Cancelling out log3 from top and bottom, we have<\/p>\n (4\/5)\/(3\/2) = (4\/5)*(2\/3) = 8\/15<\/p>\n The reason the simplification works out so cleanly is that 81 and 27 both have the same base (in this case, 3).<\/p>\n HTH:)<\/p>\n Sources:<\/p>\n Travers, Kenneth, et al. Using Advanced Algebra<\/u>. Toronto: Doubleday Canada Limited, \u00a0\u00a01977.<\/p>\n![](\"\/..\/images\/rat_exp.png\" "\\"radical")
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