(12^-x)=2.092....x=? 16^(log x / log 2)=1.43...x=?

Participant · 27 Apr 2008 at 15:38

Hello

I'm trying to solve x for the above two equations. Can someone please give me a general rule as to how to do that? Totally forgotten my Pure Maths from A-Level! I'm semi-sure with the power of -x, there's a log involved, but I just can't remember the construction.

Thanks!

Van Hellseek · 27 Apr 2008 at 15:40

Skeeter · 27 Apr 2008 at 15:43

seek said:
42

Well there goes my suggestion.

Tute · 27 Apr 2008 at 15:43

Surely the first would go to...

-xlog12 = log2.092
-x = (log2.092)/(log12)
x = -(log2.092)/(log12)

Basically the rule is.

If you have x^y = z

You can say.

ylogx = logz

And I think you can use the rule to solve the second question.