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Denebel

  • 3 years ago

How does this simplification work? Can someone show me the intermediate steps?

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  1. Denebel
    • 3 years ago
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    |dw:1346313632906:dw|

  2. juantweaver
    • 3 years ago
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    \[\ln(x)*(\frac{1}{x*\ln10})+\log(x)*(1/x)=\frac{2*\ln(x)}{x*\ln(x)}\]

  3. juantweaver
    • 3 years ago
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    Is this what you mean?

  4. Denebel
    • 3 years ago
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    Yes

  5. juantweaver
    • 3 years ago
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    okay we will need to convert log(x) to ln(x) using a given equation

  6. juantweaver
    • 3 years ago
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    we are doing this since there are no log(x) in the final answer

  7. juantweaver
    • 3 years ago
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    \[\log _{a}(x)=\frac{\ln(x)}{\ln(a)}\]

  8. juantweaver
    • 3 years ago
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    so for log base 10\[\log _{10}(x)=\frac{\ln(x)}{\ln(10)}\]

  9. juantweaver
    • 3 years ago
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    go ahead and convert this in the expression above. see what you get.

  10. Denebel
    • 3 years ago
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    |dw:1346315335776:dw|

  11. juantweaver
    • 3 years ago
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    |dw:1346315440405:dw|

  12. Denebel
    • 3 years ago
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    ^ How did you get that? Isn't whatever you do to the top, you have to do to the bottom?

  13. juantweaver
    • 3 years ago
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    you had forgotten the x in the denominator, i simply added it back

  14. juantweaver
    • 3 years ago
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    \[\frac{\ln(x)}{x*\ln(10)}+\frac{\log(x)}{x}=\frac{\ln(x)}{x*\ln(10)}+\frac{\ln(x)/\ln(10)}{x}\]

  15. juantweaver
    • 3 years ago
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    then move ln(10) to the denominator of the 2nd fraction

  16. Denebel
    • 3 years ago
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    where does the log (x) / x come from?

  17. juantweaver
    • 3 years ago
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    from the original expression that you gave me. you wrote it as \[\log(x)*\frac{1}{x}\]

  18. juantweaver
    • 3 years ago
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    \[\log(x)*\frac{1}{x}=\frac{\log(x)}{x}\]

  19. Denebel
    • 3 years ago
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    ? |dw:1346316105596:dw| from this?

  20. juantweaver
    • 3 years ago
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    yes, unless i was mistaken by what you wrote

  21. juantweaver
    • 3 years ago
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    \[\ln(x)*\frac{1}{x*\ln(10)}+\log(x)*\frac{1}{x}\] am i misunderstanding what you wrote?

  22. Denebel
    • 3 years ago
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    Ohh I see it, thank you!

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