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利用对数函数y=log_ax(a>0且a≠1)的单调性,很容易判断两个(或多个)同底对数的大小,而要判断两个异底对数的大小,却往往颇费周折。简单的,如比较log_0.30.8与log的大小,通常的解法是:第一步,作差,第二步,利用公式log_ab=1/log_ba通分,第三步,利用函数y=log_0.8x的单调性,确定分子的符号,第四步,确定分母的符号,进而确定差的符号,得出结论。拙文提出两个命题,其结论易记,易掌握,并能简化上述判断过程。 命题一:当常数a∈E(1,+∞)时,函数y=log_xa(x>0,且x≠1)(1)当且仅当0 0 and a ≠ 1), it is easy to judge the size of two (or more) logarithms of the same base. Often struggling. Simple, such as log_0.30.8 compared with the size of the log, the usual solution is: the first step, make poor, the second step, using the formula log_ab = 1 / log_ba pass points, the third step, the use of function y = log_0.8x Monotonicity, to determine the symbol of the molecule, the fourth step is to determine the denominator of the symbol, and then determine the difference of symbols, draw conclusions. I propose two propositions, the conclusion easy to remember, easy to grasp, and can simplify the process of judgment. Proposition 1: The function y = log_xa (x> 0, and x ≠ 1) when the constant a ∈ E (1, + ∞) (1) log_x_1a <0 < log_x_2a (2) In the interval (0,1) and (1, + ∞), y = loga is a reduced function. Proof (1) is the nature of the teaching function, obviously. The card (2): examine f (x_1) -f (x_2) = loga