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Sgstudent

  • 2 years ago

3 log base 5 y - 1og base 25 y = 10 This is a difficult question for me to solve and i hope you would help me thanks.

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  1. Sgstudent
    • 2 years ago
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    I know that log base 25 is actually 1/2 log base 5 y

  2. hartnn
    • 2 years ago
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    sure, first you need to know this property of log, \[\log_ba= \dfrac{\log a}{\log b}\]

  3. Sgstudent
    • 2 years ago
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    understood the property

  4. hartnn
    • 2 years ago
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    so, \(\log_5 y= ...?? \\ \log_{25}y=... ?\)

  5. Sgstudent
    • 2 years ago
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    log base 5 y = log y / log 5 and log base 25 y = log y/ log 25

  6. hartnn
    • 2 years ago
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    good :) now \(\log x^n=n \log x \\ so, \log 25=.... ?\)

  7. Sgstudent
    • 2 years ago
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    2 log 5 sir

  8. hartnn
    • 2 years ago
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    so, now you can put log y =x and the equation will become ...?

  9. Sgstudent
    • 2 years ago
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    i dont know

  10. Sgstudent
    • 2 years ago
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    can you give me the full step solution lol

  11. Sgstudent
    • 2 years ago
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    i am getting confuse here

  12. hartnn
    • 2 years ago
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    \(3x/log 5 -x/(2log 5) = 10 \\ \) got this step first ?

  13. hartnn
    • 2 years ago
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    where x=log y

  14. Sgstudent
    • 2 years ago
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    keep it going

  15. hartnn
    • 2 years ago
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    multiply both sides by 2log 5 6x - x = 20 log 5 5x = 20 log 5 got this much ?

  16. Sgstudent
    • 2 years ago
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    The first step that you posted is that the first step of showing the workings?

  17. hartnn
    • 2 years ago
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    if you understand how to solve, you will be able to show the workings on your own....

  18. Sgstudent
    • 2 years ago
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    20 log 5 means 20 log base 5?

  19. hartnn
    • 2 years ago
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    i have made the base unspecified (when its unspecified, by default its 10), so 20 log 5 means 20 times log of 5 (with the base 10)

  20. Sgstudent
    • 2 years ago
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    i dont get it

  21. Sgstudent
    • 2 years ago
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    i think that is where the problem lies in. if not, i could easily understand your workings.

  22. hartnn
    • 2 years ago
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    \(\log_ba= \dfrac{\log_{10} a}{\log_{10} b} \\\log _{25}y =\dfrac{\log_{10} y}{\log_{10} 25} \) now ?

  23. Sgstudent
    • 2 years ago
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    yes

  24. hartnn
    • 2 years ago
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    \(\log_ba= \dfrac{\log_{10} a}{\log_{10} b} \\\log _{25}y =\dfrac{\log_{10} y}{\log_{10} 25}=\dfrac{\log_{10} y}{2\log_{10} 5}\\\log_{5}y=\dfrac{\log_{10} y}{\log_{10} 5} \\ let \log y=x \\ 3x/\log 5 -x/(2\log 5)=10 \implies 6x-x=20\log_{10}5\\x=4\log_{10}5 \\ \log_{10} y = \log _{10}5^4 \implies y= 5^4\)

  25. Sgstudent
    • 2 years ago
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    okay so there is only 1 value for x?

  26. hartnn
    • 2 years ago
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    yes, and so 1 value of y.

  27. Sgstudent
    • 2 years ago
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    so what is it?

  28. Sgstudent
    • 2 years ago
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    i want to see whether i got the right answer.

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