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anonymous

  • 5 years ago

condense the expression: 1/2Log5 16-3log5 x+4Log5y

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  1. anonymous
    • 5 years ago
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    first move the coefficients that are infront of the logarithms to the right

  2. anonymous
    • 5 years ago
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    meaning log\[\log_{5} 16^{1/2}\]

  3. anonymous
    • 5 years ago
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    ignore the first log my equations button is not working

  4. anonymous
    • 5 years ago
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    but the log expression below is correct use that example to simplify the other two

  5. anonymous
    • 5 years ago
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    then let me know what you got

  6. anonymous
    • 5 years ago
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    log5y^x+4?

  7. anonymous
    • 5 years ago
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    \[\log_{5} 4-\log_{5} x ^{3}+\log_{5} y ^{4}\]

  8. anonymous
    • 5 years ago
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    thankyou for helping me their just really hard for me ;(

  9. anonymous
    • 5 years ago
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    no problem

  10. anonymous
    • 5 years ago
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    have time for more?

  11. anonymous
    • 5 years ago
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    we aren't done yet sorry, now when you are subtract two logs with the same base do you condense them by multiplying them together or dividing them?

  12. anonymous
    • 5 years ago
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    dividing

  13. anonymous
    • 5 years ago
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    yes so log base 4 is on top the numerator making the other two...?

  14. anonymous
    • 5 years ago
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    on the bottom being multiplied together

  15. anonymous
    • 5 years ago
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    idk

  16. anonymous
    • 5 years ago
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    because they were being added

  17. anonymous
    • 5 years ago
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    so can you show me what your answer would be?

  18. anonymous
    • 5 years ago
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    4/x^3+y^4

  19. anonymous
    • 5 years ago
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    when you are adding two logs together with the same base then you multiply them together

  20. anonymous
    • 5 years ago
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    okay

  21. anonymous
    • 5 years ago
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    \[\log_{5} (4/x ^{3}y ^{4})\]

  22. anonymous
    • 5 years ago
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    does that make sense

  23. anonymous
    • 5 years ago
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    yes

  24. anonymous
    • 5 years ago
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    i just want to make sure you understand rather than just giving you the answer so that later you can ace a test on this if you need to

  25. anonymous
    • 5 years ago
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    yah this good

  26. anonymous
    • 5 years ago
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    ok why we try another problem. this will be the last one i can help with, then i have to go alright?

  27. anonymous
    • 5 years ago
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    okay

  28. anonymous
    • 5 years ago
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    solve the equation; 15+2 log2 x=31

  29. anonymous
    • 5 years ago
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    so when solving for any variable the whole point is to get the variable by itself so what would be one thing we can move away from the side with x easily?

  30. anonymous
    • 5 years ago
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    yes

  31. anonymous
    • 5 years ago
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    that was a question

  32. anonymous
    • 5 years ago
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    and the answer is 15

  33. anonymous
    • 5 years ago
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    woudnt it be 17?

  34. anonymous
    • 5 years ago
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    i didn't mean the final answer, i meant that it will be easy to move the fifteen since the fifteen is not attached to anything

  35. anonymous
    • 5 years ago
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    move to be 15

  36. anonymous
    • 5 years ago
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    yes we subtract fifteen from both sides then we simplify the logarithm but first what does the new equation look like?

  37. anonymous
    • 5 years ago
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    it will equeal 16instead of 31

  38. anonymous
    • 5 years ago
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    yes

  39. anonymous
    • 5 years ago
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    know we focus on the log

  40. anonymous
    • 5 years ago
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    ;)

  41. anonymous
    • 5 years ago
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    okay!

  42. anonymous
    • 5 years ago
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    what can we move to condense the logarithm more?

  43. anonymous
    • 5 years ago
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    the 2 in front

  44. anonymous
    • 5 years ago
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    yes

  45. anonymous
    • 5 years ago
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    ah yes!!

  46. anonymous
    • 5 years ago
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    so what do we have now?

  47. anonymous
    • 5 years ago
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    logx=16

  48. anonymous
    • 5 years ago
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    no... \[\log_{2} x ^{2}=16\]

  49. anonymous
    • 5 years ago
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    does that make sense?

  50. anonymous
    • 5 years ago
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    yes

  51. anonymous
    • 5 years ago
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    okay a trick that i learned is you put the same base of the log you want to remove under both sides of the equation like so: \[2^{\log_{2}x ^{2} }=2^{16}\]

  52. anonymous
    • 5 years ago
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    now what can you simplify?

  53. anonymous
    • 5 years ago
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    exponents

  54. anonymous
    • 5 years ago
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    yes \[x ^{2}=2^{16}\] is our new equation as 2^log base 2 equals one

  55. anonymous
    • 5 years ago
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    now what would be the next step?

  56. anonymous
    • 5 years ago
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    logx=4

  57. anonymous
    • 5 years ago
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    why would we do that when we already have x by itself. what we would do is simplify 2^16=65536

  58. anonymous
    • 5 years ago
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    then x^2=65536

  59. anonymous
    • 5 years ago
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    65536?

  60. anonymous
    • 5 years ago
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    yes now what would you do to get x all by itself?

  61. anonymous
    • 5 years ago
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    take the root of x and the other side

  62. anonymous
    • 5 years ago
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    x=256

  63. anonymous
    • 5 years ago
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    does that make sense?

  64. anonymous
    • 5 years ago
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    thats what i thought

  65. anonymous
    • 5 years ago
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    \[\sqrt{x ^{2}}=\sqrt{65536}=x=256\]

  66. anonymous
    • 5 years ago
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    omg thankyou so much

  67. anonymous
    • 5 years ago
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    no problem I have to go now, but i wish you the best of luck :)

  68. anonymous
    • 5 years ago
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    thankyou

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