Question

Medal and thank you for answering this >.<
Solve for x.

12^x-2 = 3^2x+1

Answer 1

let me make sure you typed that correctly or I understand it correctly \[12^x-2 = 3^{2x}+1 \]

Answer 2

\[12^{x-2}=3^{2x+1}\]

Answer 3

the second one >.<

Answer 4

@celestialdictator

Answer 5

do you know logarithms?

Answer 6

@celestialdictator sorry I was eating >.< and I do but this problem is hard for me

Answer 7

@ChristopherToni please help!

Answer 8

http://www.mathsisfun.com/algebra/exponents-logarithms.html

Answer 9

try to study again

Answer 10

I do know the basics. I'll reread it and hopefully find a way to solve it. Thanks anyway.

Answer 11

okay !

Answer 12

@PFEH.1999 Please help. I wonder if you can give a little more in depth explanation. Thank you in advance.

Answer 13

@ramit.dour @neoh147 please help >.<

Answer 14

wait, calculating

Answer 15

\[12^{x-2}=3^{2x+1}\]
\[4^{x-2}=3^{2x+1} \div 3^{x-2}\]\[4^{x-2}=3^{x-1}\]

Answer 16

wait for next part of solution

Answer 17

Thank you both of you so, so much. <33 Much appreciated!

Answer 18

12^(x-2) = 3 ^(2x+1)
(4*3)^(x-2)=3 ^(2x+1)
4^(x-2)*3^(x-2)=3 ^(2x+1)
(4^x/4^2)*(3^x/3^2)=3^2x*3
(4^x/4^2)*(3^x/3^2)=(3^x)^2*3
divided both side with 3^x
(4^x/4^2)*(1/3^2)=3^x*3
rearrange
4^x/3^x=3^3*4^2
(4/3)^x=3^3*4^2
x= log (4/3) 3^3*4^2 4/3 is the base
change base
x= (log 3^3*4^2) / log (4/4)
use calculator....

Answer 19

sorry last line typo error....
x= (log 3^3*4^2) / log (4/3)
after press calculator,
x=21.09

Answer 20

now take log both sides
\[\ln 4^{x-2}=\ln 3^{2x+1-(x-2)}\]
(x-2)ln4=(x-1)ln3
x=(2ln4-ln3)/(-ln3+ln4)

Answer 21

x=(4ln(2)-ln(3))/(2ln(2)-ln(3)) = 5.818

Answer 22

Thank you so much for your help!! I'll rework the problem with your assistance to understand this. You guys are so awesome! 3x medals

Answer 23

http://www.wolframalpha.com/input/?i=solve+12%5E%28x-2%29+%3D+3%5E%282x%2B1%29

ans from wolfram alpha show that is 21.09...

Answer 24

@neoh147 I see, thank you so much! <3

Answer 25

you are welcome @study100