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coolaidd

solve. then round answer to the nearest hundredth. log_5x=3

  • one year ago
  • one year ago

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  1. coolaidd
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    @dpflan ?

    • one year ago
  2. dpflan
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    \[log_5x=3\]Is that the equation?

    • one year ago
  3. coolaidd
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    yess

    • one year ago
  4. dpflan
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    \[log_ab=x\] OK, this means, the number you raise a to in order to obtain x is b. so \[a^x = b\]

    • one year ago
  5. coolaidd
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    ok..

    • one year ago
  6. coolaidd
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    would it be x = 5^3 = 125?

    • one year ago
  7. dpflan
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    \[log_5x=3\] Is \[5^3 = x\]... Yeah you got it

    • one year ago
  8. coolaidd
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    what is 125 to the nearest..?

    • one year ago
  9. dpflan
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    \[125 = 125.00000...\]

    • one year ago
  10. dpflan
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    just like the last one ;)

    • one year ago
  11. coolaidd
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    what would 35 be rounded to the hundredth? 35.000?

    • one year ago
  12. coolaidd
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    ?

    • one year ago
  13. dpflan
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    Actually, no, you need one less 0. Using the decimal system, the values to the right of the decimal are fractional amounts with respect the base for the system, which is 10 here. So the first place would be \[10^{-1}\] which is 1/10, the second place is \[10^{-2}\] which is \[\frac{1}{10^2}=\frac{1}{100}\] , this that is the "hundredths" place

    • one year ago
  14. dpflan
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    so just two points to the right would be to the nearest hundrdeth

    • one year ago
  15. coolaidd
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    cool..it didnt have anything to do with the previous question..

    • one year ago
  16. coolaidd
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    i just wanted to know what 35 rounded to the nearest hundredth would be

    • one year ago
  17. dpflan
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    It's actually kind of cool, you use any number as the base. So if you have 123.456, then you have \[1*10^3 + 2*10^1 + 2*10^0 + 4*10^{-1} + 5*10^{-2} + 6*10^{-3}\]

    • one year ago
  18. dpflan
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    At least in the decimal system

    • one year ago
  19. coolaidd
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    is that for 35?

    • one year ago
  20. dpflan
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    No, 35 is 35.00

    • one year ago
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