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MTALHAHASSAN2

  • one year ago

Need help!! Solve a) log base x 625 =4

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  1. geerky42
    • one year ago
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    Hint: \(\log_x625 = \dfrac{\log 625}{\log x}\)

  2. geerky42
    • one year ago
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    Can you handle it now?

  3. Loser66
    • one year ago
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    We have equivalent formulas. Those are \(log_a b= c\iff a^c =b\)

  4. Loser66
    • one year ago
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    like \(log_2 8 = 3 \iff 2^3 =8\) Same as your problem, you can convert from log to exponent and solve for x

  5. MTALHAHASSAN2
    • one year ago
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    thnx a lot both of you

  6. MTALHAHASSAN2
    • one year ago
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    wait but how can we do this one

  7. MTALHAHASSAN2
    • one year ago
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    |dw:1432954382672:dw|

  8. MTALHAHASSAN2
    • one year ago
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    |dw:1432954474494:dw|

  9. MTALHAHASSAN2
    • one year ago
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    can someone plz help me with it

  10. MTALHAHASSAN2
    • one year ago
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    Plz

  11. anonymous
    • one year ago
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    fine

  12. anonymous
    • one year ago
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    what can I help you ? `

  13. geerky42
    • one year ago
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    How did you end up with \(\log_{x}6 = -\dfrac{1}{2}\)? Like what Loser66 said, just convert it into exponential form; \[\log_x625 = 4\quad\Longleftrightarrow\quad x^4=625\]

  14. MTALHAHASSAN2
    • one year ago
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    lol it is a different question

  15. triciaal
    • one year ago
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    |dw:1432971723597:dw|

  16. triciaal
    • one year ago
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    I wrote in exponential form then I made the power of the variable 1. maintain the balance of the equation so raised the other side to the power (-2) also rewrite with positive exponent or simple form x = 1/36

  17. MTALHAHASSAN2
    • one year ago
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    |dw:1433041180988:dw| where u get this 2 over 1 from

  18. Loser66
    • one year ago
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    |dw:1433041431862:dw|

  19. Loser66
    • one year ago
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    |dw:1433041482916:dw|

  20. Loser66
    • one year ago
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    |dw:1433041509374:dw|

  21. Loser66
    • one year ago
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    |dw:1433041552084:dw|

  22. MTALHAHASSAN2
    • one year ago
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    but why are we square rooting each side

  23. Loser66
    • one year ago
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    to get rid of 1/2

  24. triciaal
    • one year ago
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    @Loser66 wait a min

  25. Loser66
    • one year ago
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    Yes, sir

  26. triciaal
    • one year ago
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    to make the power = 1 so that the value is just the variable

  27. Loser66
    • one year ago
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    @MTALHAHASSAN2 do you agree with me that if \( 2^2 =4\) then \((2^2)^2 =4^2\)

  28. triciaal
    • one year ago
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    what am I doing or saying for someone to think I am a sir?

  29. Loser66
    • one year ago
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    hahaha... the flower shows you are a girl, right?

  30. triciaal
    • one year ago
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    2nd time in a day is a bit much

  31. triciaal
    • one year ago
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    the other person said guys like flowers too even after stating my name is pat etc

  32. triciaal
    • one year ago
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    anyway back to the question

  33. Loser66
    • one year ago
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    @triciaal we SOLVE the problem, that means we find x such that \(log_x 6=-1/2\)

  34. triciaal
    • one year ago
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    of course we did.

  35. Loser66
    • one year ago
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    in this case, we have x = 1/36, let check \(log_{1/36} 6 = -1/2?\)

  36. Loser66
    • one year ago
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    I hit my calculator, it says "you are right" hehehe

  37. MTALHAHASSAN2
    • one year ago
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    ok thnx both of you

  38. MTALHAHASSAN2
    • one year ago
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    thnx a lot

  39. Loser66
    • one year ago
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    @triciaal any question?

  40. triciaal
    • one year ago
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    @Loser66 no. I did this correctly 23 hrs ago according to this thread.

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