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coolaidd

  • 2 years ago

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

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  1. coolaidd
    • 2 years ago
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    @dpflan ?

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

  3. coolaidd
    • 2 years ago
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    yess

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

  5. coolaidd
    • 2 years ago
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    ok..

  6. coolaidd
    • 2 years ago
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    would it be x = 5^3 = 125?

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

  8. coolaidd
    • 2 years ago
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    what is 125 to the nearest..?

  9. dpflan
    • 2 years ago
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    \[125 = 125.00000...\]

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

  11. coolaidd
    • 2 years ago
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    what would 35 be rounded to the hundredth? 35.000?

  12. coolaidd
    • 2 years ago
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    ?

  13. dpflan
    • 2 years ago
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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

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

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

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

  17. dpflan
    • 2 years ago
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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 \[1*10^3 + 2*10^1 + 2*10^0 + 4*10^{-1} + 5*10^{-2} + 6*10^{-3}\]

  18. dpflan
    • 2 years ago
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    At least in the decimal system

  19. coolaidd
    • 2 years ago
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    is that for 35?

  20. dpflan
    • 2 years ago
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    No, 35 is 35.00

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