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anonymous

  • 5 years ago

Hi, me again. Can anybody help me solve [ 5^(x-4) = 6^(2x+1) ] with natural logarithms?

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  1. anonymous
    • 5 years ago
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    Take logarithms on both sides of equality: It will be \[(x-4)\log5=(2x+1)\log6 \implies x=\frac{\log6+4\log5}{\log5-2\log6}\]

  2. anonymous
    • 5 years ago
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    Aweome! Thanks so much for your help tonight; my teacher's given us very sparse information on the topic and none of the practice questions have answers with working. Thanks again!

  3. anonymous
    • 5 years ago
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    Hmm, just to make sure, exactly how did you rearrange it to get that final fraction?

  4. anonymous
    • 5 years ago
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    Multiply x with log5 and then -4 with log5. Its just the distributive property. \[(x-4)\log5=x\log5-4\log5 \ and\ (2x+1)\log6=2x\log6+\log6\]

  5. anonymous
    • 5 years ago
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    then equate the two sides collecting the x terms and the constant terms. Its easy right?

  6. anonymous
    • 5 years ago
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    Oh, like indices! Yep, thanks saubhik! :)

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