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    Hi,

    I've got a log question which I've solved using the method I'm most familiar with but the question wants it in a specific form that I'm struggling to get to.

    The question is: given that  2^{3x-5} = 3^{3-7x} , show that x= \frac{3k+5}{7k+3} , k=\frac{ln3}{ln2} .

    So far I've got it as  x=\frac{3ln3 + 5ln2}{3ln2+7ln3} but can't get passed that! Is there some really obvious log law that I've failed to consider?!
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    (Original post by marcsaccount)
    Hi,

    I've got a log question which I've solved using the method I'm most familiar with but the question wants it in a specific form that I'm struggling to get to.

    The question is: given that  2^{3x-5} = 3^{3-7x} , show that x= \frac{3k+5}{7k+3} , k=\frac{ln3}{ln2} .

    So far I've got it as  x=\frac{3ln3 + 5ln2}{3ln2+7ln3} but can't get passed that! Is there some really obvious log law that I've failed to consider?!
    No, but you could divide the top and bottom of your fraction by ln2 and then compare with what they want you to get
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    No law.

    If you log both sides, you should get

    (3x-5)ln2= (3-7x)ln3

    The next step should be to divide both sides by ln2 and then sub k=ln3/ln2

    Take it from there.
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    (Original post by davros)
    No, but you could divide the top and bottom of your fraction by ln2 and then compare with what they want you to get

    Ahhhh yeah! Cheers Davros!
 
 
 
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