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Q1 - Equation with fractional and logarithmic exponents

  1. Aug 23, 2010 #1
    1. The problem statement, all variables and given/known data

    4 1/(x+1) · 8 1/(x+2) = 9log3 2

    2. Relevant equations

    n/a - same as 1

    3. The attempt at a solution

    To get rid of the exponents and be able to solve for x, I first took the logarithm with base 4 of all terms:

    log44 1/(x+1) · log48 1/(x+2) = log49 log3 2

    This equals:

    1/(x+1) · 1/(x+2) · log48 = log32 · log49

    When adding the fractions the left side equals:

    log48/((x+1)(x+2))

    Here I already feel I am on the wrong track, but it is still possible to get rid of the fraction by multiplying by the LCM (x+1)(x+2):

    log4 8 = (x+1)(x+2)log32 · log49

    Now it is possible to get the x's on one side by dividing both sides by log32 · log49:

    log48/(log32 · log49) = (x+1)(x+2)

    The x's can be worked free as: x2+3x+2, thus:

    log48/(log32 · log49) -2 = x2+3x

    The right side can now be made to look like: x(x+3), however, this does not solve x!

    Now I am at a loss here, I have the feeling the three exponents at the start should be freed much sooner and in a cleaner way then I tried above, but this is the closest I came to a solution for x. Some help would be appreciated!

    Thanks in advance,
     
  2. jcsd
  3. Aug 23, 2010 #2

    eumyang

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    You can simplify the right side before you even take the log base 4 of both sides. Note that
    [tex]b^{\log_b n} = n[/tex], so

    [tex]\begin{aligned}
    9^{\log_3 2} &= (3^2)^{\log_3 2} \\
    &= 3^{2 \log_3 2} \\
    &= 3^{\log_3 4} \\
    &= 4
    \end{aligned}[/tex]

    Now take the log base 4 of both sides. When you did that originally, I noticed that on the left side you had this:
    log44 1/(x+1) · log48 1/(x+2)

    You should use a single log for the product, instead of a log for each factor, like this:
    [tex]\log_4 \left( 4^{1/(x + 1)} \cdot 8^{1/(x + 2)} \right)[/tex]

    Furthermore, [tex]\log_4 8[/tex] can be simplifed to a rational number:
    [tex]\log_4 8 = \log_4 4^{3/2} = \frac{3}{2}[/tex]


    69
     
    Last edited: Aug 23, 2010
  4. Aug 23, 2010 #3

    ehild

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    Gold Member

    Wrong. log(a*b)=log(a)+log(b).

    ehild
     
  5. Aug 23, 2010 #4

    hunt_mat

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    Can't you write 4=2^2, and 8=2^3 to obtain:
    [tex]
    2^{2/(1+x)}\cdot 2^{3/(x+2)}=3^{4\log 3}
    [/tex]
    Which when rearranged shows:
    [tex]
    2^{2/(1+x)+3/(x+2)}=3^{4\log 3}
    [/tex]
    From here it looks straight forward to solve.

    Mat
     
  6. Aug 23, 2010 #5
    Thank you all for the quick responses.

    I have one more quick question about taking the logarithm of a side of an equation.

    Consider the equation

    32x+1 - 28 * 3x + 9 = 0

    Now, it is obvious that a logarithm with base 3 needs to be taken, what I am wondering about is whether I have distributed the signs appropriately in the following calculation:

    log332x+1 - (log328 + log33x) + log39 = 0

    implies

    2x + 1 - log328 - x + 2 = 0

    implies the solution

    x = log328 - 3
     
  7. Aug 23, 2010 #6

    hunt_mat

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    write u=3^{x}, then your equation reduces to:
    [tex]
    3u^{2}-28u+9=0
    [/tex]
    Solve this and then take logs to base 3.
     
  8. Aug 23, 2010 #7
    I don't seem to follow you completely.

    If 3x = u

    How is 32x+1 = 3u2 ??
     
  9. Aug 23, 2010 #8

    hunt_mat

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    This use nothing but the laws of powers.
    [tex]
    (3^{x})^{2}=3^{2\cdot x}=3^{2x}
    [/tex]
    Then:
    [tex]
    3\cdot 3^{2x}=3^{2x+1}
    [/tex]
     
  10. Aug 23, 2010 #9
    Ok got it...wow, I really should learn to take all rules into account, I find it hard to see those things straight away like you seem to be able to :S

    Here is one last one:

    lg(x2-5x+7) / lg(4-x) = 0

    I assume with lg log10 is meant.

    Because this is a fraction both the numerater and denominator may be multiplied by the same number, in this case:

    log1010(x2-5x+7)/log1010(4-x)

    Is this correct?
     
  11. Aug 23, 2010 #10

    hunt_mat

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    You have an equation:
    [tex]
    \frac{\log(x^{2}-5x+7)}{\log (4-x)}=0
    [/tex]
    Multiply though by log (4-x), to obtain:
    [tex]
    \log (x^{2}-5x+7)=0
    [/tex]
    So this means that:
    [tex]
    x^{2}-5x+7=1\Rightarrow x^{2}-5x+6=0
    [/tex]
     
  12. Aug 23, 2010 #11

    hunt_mat

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    I am teaching this sortof stuff at the moment, so don't feel too bad.
     
  13. Aug 23, 2010 #12
    Allright, one more and I should have a clear understanding of everything they can throw at me (hopefully).

    32+log925 + 251-log52 + 10-lg4

    Now, From the previous problem, I have a feeling that before doing anything to all of the terms, first the terms have to be simplified individually.

    For the last one I can see that:

    10-lg4 = 10lg4-1 = 4-1 = 1/4

    For

    32+log925

    I can see there is a 32 in there somewhere, but how to get it out of the 2+log925?

    Is

    32(1+.5log925) = 91+.5log925

    A valid way of doing this? If so, how to continue from here?

    For

    251-log52 = 52(1-log52)

    And if I continue

    52-2log52 = 52-log54 = 2-4 ???

    What to do with those numbers that are not part of the logarithm?
     
  14. Aug 23, 2010 #13
    Is replacing 2 with the logarithm with base 5 that equals 2 a solution?

    5log525-log54 = 5log5(25/4) = 25/4???
     
  15. Aug 23, 2010 #14

    hunt_mat

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    Sort of, I can't quite read your equations, can you make then a little clearer please.
     
  16. Aug 23, 2010 #15
    Hmm and for

    32+log925 = 32(1+.5log925) = 91+log95

    This would mean:

    9log99+log95 = 9log9(9*5) = 45

    Please tell me I'm right...
     
  17. Aug 23, 2010 #16

    eumyang

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    Note that
    [tex]a^{m + n} = a^m \cdot a^n[/tex]
    and
    [tex]b\log a = \log a^b[/tex]

    You'll need these rules to simplify further.

    This is not correct. Use this rule:
    [tex]a^{m - n} = \frac{a^m}{a^n}[/tex]


    69
     
    Last edited: Aug 23, 2010
  18. Aug 23, 2010 #17
    Ok I hope this is clearer:

    10^-lg 4 = 10^lg 4-1 = 4-1 = 1/4

    3^2+log925 = 3^2(1+.5 log925) = 9^1+log95 = 9^log99+log95 = 9^log9(9*5) = 45
     
  19. Aug 23, 2010 #18
    And for the last:

    25^1-log52 = 5^2(1-log52) = 5^2-2 log52 = 5^2-log54 = 5^log525-log54 = 5^log5(25/4) = 25/4
     
  20. Aug 23, 2010 #19

    eumyang

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    Actually, not really. I'd learn LaTeX if I were you.
    Or,
    [tex]3^{2 + \log_9 25} = 3^{2(1 + 0.5\log_9 25)} = 9^{1 + 0.5\log_9 25} = 9^1 \cdot 9^{0.5\log_9 25} = 9 \cdot 9^{\log_9 5} = 9 \cdot 5 = 45[/tex]


    69
     
  21. Aug 23, 2010 #20
    Well, I'd rather spend time studying these problems at the moment, I have an exam to pass on friday...

    Anyway, is it difficult? I mean if it's easy I might as well use it.

    Anyhow, the last "translation" you did was what I wrote...is it correct?
     
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