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anonymous

  • 5 years ago

Simplify log(base a)1-log(base a)a^b

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  1. anonymous
    • 5 years ago
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    log(1/(a^b)) => log(a^(-b)) => -blog(a) => -b sing base is a we have -b

  2. anonymous
    • 5 years ago
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    Remember that by definition, for any number x, $$\log_x 1= 0,$$ since any number to the zeroth power is 1. Also, remember that, by definition for any number x, $$\log_x x^b = b.$$ That's what a logarithm is. So combining those two: $$\log_a 1 - \log_a a^b = 1 - b$$ That's right, the a doesn't matter at all!

  3. anonymous
    • 5 years ago
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    Sorry, I that last equation was supposed to read: $$\log_a 1 - \log_a a^b = -b$$

  4. anonymous
    • 5 years ago
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    Rule #1 of posting corrections: your correction will itself contain a mistake. <goes away grumbling darkly/>

  5. anonymous
    • 5 years ago
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    thanks, i couldnt put the equation thing. so the answer is just -b??? that simple?

  6. amistre64
    • 5 years ago
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    yes, -b

  7. anonymous
    • 5 years ago
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    but you said logx1=0 so shoudltn it be 0-b?

  8. amistre64
    • 5 years ago
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    0 is not a "simplification"; it is an overcomplication :)

  9. amistre64
    • 5 years ago
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    0 -b = -b ; see :)

  10. anonymous
    • 5 years ago
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    ok :) thanks

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