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

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

Mathematics
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Take logarithms on both sides of equality: It will be \[(x-4)\log5=(2x+1)\log6 \implies x=\frac{\log6+4\log5}{\log5-2\log6}\]
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!
Hmm, just to make sure, exactly how did you rearrange it to get that final fraction?

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Multiply x with log5 and then -4 with log5. Its just the distributive property. \[(x-4)\log5=x\log5-4\log5 \ and\ (2x+1)\log6=2x\log6+\log6\]
then equate the two sides collecting the x terms and the constant terms. Its easy right?
Oh, like indices! Yep, thanks saubhik! :)

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