Difference between revisions of "TDSM 2.15"
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(Created page with "The derivative of <math>f(x) = \log_a x</math> (<math>x > 0</math>) is <math>\frac{\ln{a}}{x} > 0</math> <math>\forall x > 0</math> so <math>\log_a x</math> is an increasing f...") |
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| − | The derivative of <math>f(x) = \log_a x</math> (<math>x > 0</math>) is <math>\frac{\ln | + | The derivative of <math>f(x) = \log_a x</math> (<math>x > 0</math>) is <math>\frac{1}{x\ln(a)} > 0</math> <math>\forall x > 0</math> so <math>\log_a x</math> is an increasing function. |
With <math>x < 1</math> we have <math>f(x) < f(1) = \log_a 1 = 0</math> (proved) | With <math>x < 1</math> we have <math>f(x) < f(1) = \log_a 1 = 0</math> (proved) | ||
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| + | --- | ||
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| + | However, when 0 < a < 1, <math>\log_a x</math> is a ''decreasing'' function. it is not true. | ||
Latest revision as of 22:11, 12 December 2017
The derivative of [math]f(x) = \log_a x[/math] ([math]x \gt 0[/math]) is [math]\frac{1}{x\ln(a)} \gt 0[/math] [math]\forall x \gt 0[/math] so [math]\log_a x[/math] is an increasing function.
With [math]x \lt 1[/math] we have [math]f(x) \lt f(1) = \log_a 1 = 0[/math] (proved)
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However, when 0 < a < 1, [math]\log_a x[/math] is a decreasing function. it is not true.
