Question

Solve exactly pi^(-2x+1)=e^x

Answer 1

\(\bf \pi^{-2x+1}=e^x\qquad \textit{taking "ln" to both sides}\\ \quad \\
log_e(\pi^{-2x+1})=log_e(e^x)\\ \quad \\
\textit{log cancellation rule of }\quad log_aa^x=x\qquad thus\\ \quad \\
log_e(\pi^{-2x+1})=log_e(e^x)\implies log_e(\pi^{-2x+1})=x\)

... so... what do you think about the left side? can we ... do something with the exponent? maybe a log exponent rule?

Answer 2

\(\Large \bf \textit{recall that}\quad log_a(x^y)\implies ylog_ax\)

Answer 3

... so... what do you think?

Answer 4

@jdoe0001 is always there to solve problems I find interesting. :P

Answer 5

Thank you! So now, can we change it into exponential form, so that, e^x=-2x+1, or does that not help?

Answer 6

hehe

Answer 7

\(\bf \pi^{-2x+1}=e^x\qquad \textit{taking "ln" to both sides}\\ \quad \\

log_e(\pi^{-2x+1})=log_e(e^x)\implies log_e(\pi^{-2x+1})=x\\ \quad \\
(-2x+1)log_e(\pi)=x\) so if you distribute the left-side,... what would it look like?

Answer 8

\(\bf \pi^{-2x+1}=e^x\qquad \textit{taking "ln" to both sides}\\ \quad \\

log_e(\pi^{-2x+1})=log_e(e^x)\implies log_e(\pi^{-2x+1})=x\\ \quad \\
(-2x+1)log_e(\pi)=x\)

so if you distribute the left-side,... what would it look like?

Answer 9

What is it that we are distributing here?

Answer 10

\(\bf {\Large (-2x+1)log_e(\pi)}=x\)

Answer 11

So if I plug in ln for loge I would have (-2x+1)lnpi=x ? Which would then be ln(-2x)+ln+lnpi ?

Answer 12

well... you can think of it as say .... \(\bf (-2x+1)log_e(\pi)=x\qquad log_e(\pi)=a\implies (-2x+1)a=x\) if you distribute that... what would it give you?

Answer 13

Oh okay, -2xa+a=x

Answer 14

yeap.. so \(\bf (-2x+1)log_e(\pi)=x\qquad log_e(\pi)=a\implies (-2x+1)a=x\\ \quad \\
\implies -2xa+a\implies -2xlog_e(\pi)+log_e(\pi)\qquad thus\\ \quad \\

(-2x+1)log_e(\pi)=x\implies (-2x)log_e(\pi)+log_e(\pi)=x\\ \quad \\
(-2x)log_e(\pi)+log_e(\pi)-x=0\implies (-2x)log_e(\pi)-x=-log_e(\pi)\\ \quad \\
\textit{now taking common factor to the left-side}\\ \quad \\
x[]=\implies x=\cfrac{-log_e(\pi)}{-2log_e(\pi)-1}\)

Answer 15

hmm hhehhe.l missing a piece.... there... ok

Answer 16

\(\bf (-2x+1)log_e(\pi)=x\qquad log_e(\pi)=a\implies (-2x+1)a=x\\ \quad \\
\implies -2xa+a\implies -2xlog_e(\pi)+log_e(\pi)\qquad thus\\ \quad \\

(-2x+1)log_e(\pi)=x\implies (-2x)log_e(\pi)+log_e(\pi)=x\\ \quad \\
(-2x)log_e(\pi)+log_e(\pi)-x=0\implies (-2x)log_e(\pi)-x=-log_e(\pi)\\ \quad \\
\textit{now taking common factor to the left-side}\\ \quad \\
x[-2log_e(\pi)-1]=-log_e(\pi)\implies x=\cfrac{-log_e(\pi)}{-2log_e(\pi)-1}\)

Answer 17

Great! So then, is it acceptable to cancel out the negative signs which would result in ln(pi)/(1+2ln(pi))...?

Answer 18

sure, yes

Answer 19

\[\ln(\pi)/ (1+2\ln(\pi))\]

Answer 20

yeap

Answer 21

Great, thank you so much!

Answer 22

yw