reply on: [Daily puzzle] Simple algebra problem. \ stacker news ~science

pull down to refresh

Not sure how you got to this expression...

 $\frac{(27*343)^x}{(63*147)^x} = \frac{37}{21}$

 $(\frac{27*343}{63*147})^x= \frac{37}{21}$

 $1^x = \frac{37}{21}$

I can't remember how to solve x. (log?)

 $a^x + b^x \neq (ab)^x$

You can easily check this inequality using

a=2

b=3

and

x=2

I've been looking at it wrong. Where do you have +, I saw x

 $a^x \times b^x = (ab)^x$

I need to replace the lenses in my glasses

Honest mistake ;)