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anonymous

  • 5 years ago

how do u solve log of x 5=1/4

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  1. anonymous
    • 5 years ago
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    You need to be a little clearer. Is it,\[\log_x5=\frac{1}{4}\]or\[\log x^5=1/4\]?

  2. anonymous
    • 5 years ago
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    1st

  3. anonymous
    • 5 years ago
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    ok

  4. anonymous
    • 5 years ago
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    one sec...dealing with a couple of things at once.

  5. anonymous
    • 5 years ago
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    You have to use the definition of what the logarithm means. If you have\[y=a^x\]then by definition of the logarithm,\[\log _a y = x\] By the definition then, you have\[\log_x 5 = 1/4 \rightarrow x^{1/4}=5\]Raise both sides to the power of 4, and you have\[(x^{1/4})^4=5^4 \rightarrow x=625\]

  6. anonymous
    • 5 years ago
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    i have another u could help me with

  7. anonymous
    • 5 years ago
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    and thanks

  8. anonymous
    • 5 years ago
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    ok baseballkid, but I could use another fan :p

  9. anonymous
    • 5 years ago
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    ln(5x-3)+ln2- ln(24-2x)

  10. anonymous
    • 5 years ago
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    how do i become your fan

  11. anonymous
    • 5 years ago
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    You have to know your log laws. The two you need here are:\[\log a + \log b = \log ab\]and\[\log a - \log b = \log \frac{a}{b}\]Here, for the frist two terms, you have\[\ln ( 5x-3)+\ln 2 = \ln (2(5x-3))=\ln (10x-6)\]and then\[\ln (10x-6) -\ln (24-2x)= \ln \frac{10x-6}{24-2x}=\ln \frac{5x-3}{12-x}\]since the numerator and denominator have a common factor of 2.

  12. anonymous
    • 5 years ago
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    ok im an idiot i put it in wrong i meant to put ln(5x-3) + ln2 = ln(24-2x)

  13. anonymous
    • 5 years ago
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    but why r u so smart? i mean i think its great

  14. anonymous
    • 5 years ago
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    Okay. Now you're solving for x...that's different.

  15. anonymous
    • 5 years ago
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    lol, so smart...thanks...

  16. anonymous
    • 5 years ago
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    You need to exponentiate both sides of your equation:\[e^{\ln(5x-3)+\ln2}=e^{\ln(24-2x)} \rightarrow e^{\ln2(5x-3)}=e^{\ln(24-2x)}\]\[\rightarrow 2(5x-3)=24-2x\]

  17. anonymous
    • 5 years ago
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    You expand and solve for x now.

  18. anonymous
    • 5 years ago
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    Do you see what I did with exponentiation? It 'undoes' the logarithm. It's the 'inverse' of logarithm.

  19. anonymous
    • 5 years ago
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    Once you are done with this, could you please provide me some help

  20. anonymous
    • 5 years ago
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    I may have to just take a look and go. I'm supposed to be elsewhere today.

  21. anonymous
    • 5 years ago
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    Ok it will only take a look from a genius like you Here it is http://openstudy.com/groups/mathematics#/updates/4da834d1d6938b0bf8d5a44d

  22. anonymous
    • 5 years ago
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    baseballkid, are you okay with this? If not, leave a message here and when I can, I'll get back to you.

  23. anonymous
    • 5 years ago
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    ok thank you so much. i have some more ill post later

  24. anonymous
    • 5 years ago
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    ok

  25. anonymous
    • 5 years ago
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    post them for others in the question box if you need them quickly (others will do them), else, post here and I'll do what I can. :)

  26. anonymous
    • 5 years ago
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    \[\log_{5} (3x+10)-3\log_{5}4=2 \]

  27. anonymous
    • 5 years ago
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    You have to use the log laws,\[\log_5 (3x+10)-3\log _ 5 4 =\log_5 (3x+10)-\log _ 5 4 ^3\]\[=\log 5 \frac{3x+10}{64}\]that is,\[\log _ 5 \frac{3x+10}{64}=2 \rightarrow \frac{3x+10}{64}=5^2\]by definition of the logarithm. Solving for x, you have\[3x+10=25 \times 64 \rightarrow x = 530\]

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