anonymous 5 years ago How could I find log4y+log2y=12?

1. anonymous

rasing e to the relative powers gives $e ^{\log(4y)+\log(2y)}=e ^{12}$ seperating the powers of e then gives $e ^{\log(4y)}e ^{\log(2y)} = e ^{12}$ hence $8y ^{2} = e ^{12}$ so solving from there gives $y=e^{6}/2\sqrt{2}$

2. anonymous

agreed

3. anonymous

Thank alot aclandt, I was stuck in this question for more than 5hrs

4. anonymous

assuming log with base e

5. anonymous

yeah, though unless given otherwise you should always assume log to be to base e