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

Question

anonymous · Answer

Take logarithms on both sides of equality:
It will be $$(x-4)\log5=(2x+1)\log6 \implies x=\frac{\log6+4\log5}{\log5-2\log6}$$

anonymous · Answer

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!

anonymous · Answer

Hmm, just to make sure, exactly how did you rearrange it to get that final fraction?

anonymous · Answer

Multiply x with log5 and then -4 with log5. Its just the distributive property.
$$(x-4)\log5=x\log5-4\log5 \ and\ (2x+1)\log6=2x\log6+\log6$$

anonymous · Answer

then equate the two sides collecting the x terms and the constant terms. Its easy right?

anonymous · Answer

Oh, like indices! Yep, thanks saubhik! :)