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

1. anonymous

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

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

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

4. anonymous

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

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

6. anonymous

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