what am i doing wrong?!!? on this log. problem ..?

Question

okay .. its PI^1-x = e^x

then i do ..

1-x ln PI = x ln e
1-x ln PI = x

.. and it goes all downhill from there :( .. what to do next and why? i usually get stuck right here on all the problems .. what is the property i need to know?

Answer 1

Are you solving for x?

pi^(1-x) = e^x
(1-x)ln pi = x
ln pi - xln pi = x
ln pi = x + xln pi
ln pi = x(1 + ln pi)
ln pi/(1 + ln pi) = x

You almost had it! You seem to understand your logs, don't forget to go back to the basics now and then!

Answer 2

ok... here's what you do...

you did good so far. You need to gather up all the x's on the same side of the equal sign, so here's what I'd do from the last step.

(1-x) ln pi = x
ln pi - x ln pi = x
ln pi = x + x ln pi
ln pi = x(1 + ln pi)
(ln pi) / (1 + ln pi) = x

so there ya go...

x = (ln pi) / (1 + ln pi)

Answer 3

First let´s write it right
Pl^(1-x)=e^x
now take ln
(1-x)lnPL =x Apply distributive rule
Pl= x(1+lnPl)
so x= lnPl/(1+lnPl)

Answer 4

pi^(1 - x) = e^x
(1 - x)ln(pi) = xln(e)
ln(pi) - xln(pi) = x
-x - xln(pi) = ln(pi)
-x(1 + ln(pi)) = -ln(pi)
x(1 + ln(pi)) = ln(pi)
x = ln(pi)/(1 + ln(pi))

all you needed to do is multiply ln(pi) by (1 - x), then solve for x.