Question

solve: 2^(x+1) = 3^(1-2x)

Answer 1

\[log_e both sides \]
\[xlog_e2 + log_e2 =log_e3 -2xlog_e3\]
Solve this

Answer 2

\[\ln(2^{x+1})=\ln(3^{1-2x})\]
\[(x+1)\ln2=(1-2x)\ln3\]
\[xln2+\ln2=\ln3-2xln3\]
\[x(\ln2+2\ln3)=\ln3-\ln2\]
\[x=\frac{\ln3-\ln2}{\ln2+2\ln3}\]

Answer 3

\[(x+1)\ln(2)=(1-2x)\ln(3)\] and now a bunch of algebra

Answer 4

\[x\ln(2)+\ln(2)=\ln(3)-2x\ln(3)\]

Answer 5

and now look above!

Answer 6

\[x=\frac{\ln(\frac{3}{2})}{\ln2+\ln9}=\frac{\ln(\frac{3}{2})}{\ln(2 \cdot 9)}=\frac{\ln(\frac{3}{2})}{\ln18}\]

Answer 7

myininaya, are you the type that writes
\[\ln3\] and
\[\sin x\]?

Answer 8

?

Answer 9

i will report you to the math police

Answer 10

\[\sin(x),\ln(x)\]

Answer 11

i do it both ways

Answer 12

\[\frac{\ln6}{\ln3}=2\]

Answer 13

yes?

Answer 14

no

Answer 15

why not? cancel the 3, cancel the ln

Answer 16

\[\frac{\ln9}{\ln3}=2\]

Answer 17

no
\[\frac{\ln 9}{\ln3}=3\]

Answer 18

satellite why you always hating on me

Answer 19

lol

Answer 20

maybe, but if i have
\[\frac{\ln(9)}{\ln(3)}\] then i know it is 2. otherwise i just cancel the ln and the 3 to get 3

Answer 21

i like to give you a hard time, i will stop.
ok i won't

Answer 22

"why you be hatin'?"

Answer 23

ln by itself makes no sense
clearly when i say ln3
it also means ln(3)

Answer 24

and when i say eggs i mean bacon

Answer 25

no!

Answer 26

LOL