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Homework Help: Logarithms T_T

  1. Jul 11, 2007 #1



    Hi

    I really need ur help

    I was sick in the days that the teacher gave us the lesson

    and I really really need ur help



    this is the equation :


    2 log{to base x} of 5 + log{to base 5} of x = log 1000


    i know that log 1000 = 3


    and 2 log{to base x} of 5 = log{to base x} of 25


    so

    log{to base x} of 25 + log{to base 5} of x = 3

    and now I don't know how to continue :confused:

    cause I don't know how to add different bases :grumpy:



    plz someone helps me

    how can I continue and how can I add different bases ?!

    HELP ME :cry: ..



     
  2. jcsd
  3. Jul 11, 2007 #2

    Dick

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    Don't add different bases. Convert the base. Do you know:

    log{base a}b=log{base c}b/log{base c}a.

    Where c is anything. Use this to convert log{base x}25 to an expression involving log{base 5} only. I.e. put c=5 in the above.
     
  4. Jul 12, 2007 #3

    VietDao29

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    Yup, so far, so good. :)

    Since you cannot add different bases. You need to convert it to only one base. The Change of Base Formula would did it:

    [tex]\log_{a}b = \frac{\log_c b}{\log_c a}[/tex]

    This is what Dick told you.

    From the above Formula, we can derive another 2 pretty nice, and also important Formulae:
    1. [tex]\log_a b \times \log_b c = \log_a c[/tex]

    2. [tex]\log_a b = \frac{\log_b b}{\log_b a} = \frac{1}{\log_b a}[/tex]

    Remember the three formulae above, you may need it in Logarithmic Problems. :)
     
  5. Jul 12, 2007 #4


    ok i think i got it

    but don't laugh if it's wrong

    2 log {base x} of 5 + log {base 5} of x = log 1000

    1/( 2 log {base 5} of x ) + ( log {base 5} of x )/1 = 3

    1/( 2 log {base 5} of x ) + (( log {base 5} of x )( 2 log {base 5} of x ))/( 2 log {base 5} of x ) = 3

    so

    ( log {base 5} of x )/( 2 log {base 5} of x ) + 1 = 3

    ( log {base 5} of x )/( 2 log {base 5} of x ) = 3 -1

    ( log {base 5} of x )/( 2 log {base 5} of x ) =2

    1/x = 2

    so

    x=1/2


    am i right ?!

    :shy:

     
  6. Jul 12, 2007 #5

    Dick

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    Nope. Not quite. Let's call L=log{base 5}x, then yes, log{base x}5=1/L. But compare your first and second equations. How did the 2 move from the numerator to the denominator?
     
  7. Jul 12, 2007 #6


    ok so u'r saying that what i did was correct

    that 1/x = 2

    but x doesn't equal 1/2

    right ?!

     
  8. Jul 12, 2007 #7

    Dick

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    No, I'm saying there are problems near the start. Your second equation is 1/(2L)+L=3 (again, let L=log{base 5}x). I'm trying to convince you that it should be 2/L+L=3.
     
  9. Jul 12, 2007 #8



    ok so

    2 log {base x} of 5 + log {base 5} of x = log 1000

    2 / 2 log {base 5} of x + log {base 5} of x = 3

    2 / log {base 5} of x^2 = 3

    2 / 2 log {base 5} of x = 3

    1 / log {base 5} of x = 3

    log {base x} of 5 = 3

    right ?!

    :biggrin:


     
  10. Jul 12, 2007 #9

    Dick

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    Still wrong. You are doing some really funny looking algebra. How does

    2 log{base x}5 become 2/(2 log{base 5}x)? It should be just

    2/(log{base 5}x). Where is the extra 2 coming from?
     
  11. Jul 12, 2007 #10


    ok i think i got it

    2 log {base x} of 5 + log {base 5} of x = log 1000

    [2/ log {base 5} of x] + log {base 5} of x = 3

    ok then i should

    ummm ..

    [2 (log {base 5} of x) (log {base 5} of x)] / log {base 5} of x =3

    2 (log {base 5} of x) =3

    so

    (log {base 5} of x^2) = 3

    am i right this time ?!

    :biggrin:

     
  12. Jul 12, 2007 #11

    Dick

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    Not right yet, but making progress. You now have 2/L+L=3. (Where L is your log{base 5}x. Can you solve 2/L+L=3 for L? I think the wordiness of your notation may be confusing you. Solve 2/L+L=3. It's the same equation. What's the first step?
     
  13. Jul 12, 2007 #12


    cool i'm making progress

    ok

    2/L+L=3

    first we multiply both sides by 1/2

    so it will be

    L+L = 3/2

    2L = 3/2

    so

    L = 3/4


    am i right ?!

    :shy:

     
  14. Jul 12, 2007 #13

    Dick

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    Sorry, but I think you need some serious practice on basic algebra.

    (1/2)*(2/L+L)=(1/2)*(2/L)+(1/2)*L=1/L+L/2. That's the correct result - but it didn't simplify things much. What you really want to do is multiply both sides by L (to get rid of the L in the denominator). You are going to get a quadratic equation. Can you handle those?
     
  15. Jul 12, 2007 #14



    :frown:

    i think the teacher was telling me the truth when she said i'm stupid

    2 log {base x} of 5 + log {base 5} of x = log 1000

    [2/ log {base 5} of x] + log {base 5} of x = 3

    L = log {base 5} of x

    2/L + L = 3

    2+L^2 = 3L

    L^2 - 3L + 2 = 0

    ( L-2 ) ( L -1 ) = 0

    L = 2 or L = 1

    ----------------

    what do i do now ?!


     
  16. Jul 12, 2007 #15

    Dick

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    Now you are ready to collect your prize. L=log{base 5}x=1 or 2. Can you solve for x in each of those two cases?
     
  17. Jul 12, 2007 #16


    ok

    if

    log {base 5} of x =1

    then x = 5

    and if

    log {base 5} of x = 2

    then x = 25

    am i right ?!

    :shy:

     
  18. Jul 12, 2007 #17

    Dick

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    You win!!!! Yes. Can you check that they both work in the original equation? I was serious about doing some algebra practice if you want to tackle these questions on your own.
     
  19. Jul 12, 2007 #18


    thank u

    thank u

    thank u

    i can't believe that i solved it

    actually i couldn't solve it without u

    thank u

    thank u

    thank u


    ok and now i'll check

    2 log {base x} of 5 + log {base 5} of x = log 1000

    2 log {base 5} of 5 + log {base 5} of 5 = 3

    2 * 1 + 1 = 3

    2 + 1 = 3

    :biggrin:

    ========

    2 log {base x} of 5 + log {base 5} of x = log 1000

    2 log {base 25} of 5 + log {base 5} of 25 = 3

    2 * 0.5 + 2 = 3

    1 + 2 = 3

    :biggrin:

    ========

    i'm so happy :biggrin: :biggrin: :biggrin:

    and again

    thank u

    thank u

    thank u

     
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