# Simplify this equation.

• Oct 22nd 2007, 02:22 PM
Takkun
Simplify this equation.
y=2e^-(ln x) + 1

Hey, can someone help me simplify this equation.
• Oct 22nd 2007, 02:25 PM
TD!
You know that exp(ln(a)) = a? And r.ln(x) = ln(x^r), so -ln(x) = ln(1/x).
• Oct 22nd 2007, 02:27 PM
Takkun
Quote:

Originally Posted by TD!
You know that exp(ln(a)) = a? And r.ln(x) = ln(x^r), so -ln(x) = ln(1/x).

so...

y=2(1/x)+1
y=2/x +1
?
• Oct 22nd 2007, 02:29 PM
topsquark
Quote:

Originally Posted by Takkun
y=2e^-(ln x) + 1

Hey, can someone help me simplify this equation.

Or you could look at it this way:
$\displaystyle y = 2e^{-ln(x)} + 1 = \frac{2}{e^{ln(x)}} + 1$
and then use TD's trick: $\displaystyle e^{ln(a)} = a$.

(It isn't much different, but I prefer to have my exponents positive when possible.)

-Dan
• Oct 22nd 2007, 02:30 PM
TD!
Quote:

Originally Posted by Takkun
so...

y=2(1/x)+1
y=2/x +1
?

Correct! :)
• Oct 22nd 2007, 02:30 PM
topsquark
Quote:

Originally Posted by Takkun
so...

y=2(1/x)+1
y=2/x +1
?

Yes, that's it. :)

-Dan