help?!!? logarithms!?

Question

3^1-2x = 4^x


.. i think the next step is 1-2x ln 3 = x ln 4
.. is that right? what do i do next??

the answer is .. ln 3/ 2 ln3 + ln4

You should put parenthesis
3^(1-2x)=4^x
(1-2x)ln3 =xln4
x(2ln3+ln4)= ln3 so
x=ln3/(2ln3+ln4)
.....

you lost me .. how did you get x(2n3+ln4)=ln3 ???

Answer 1

3^(1-2x) = 4^x
ln 3^(1-2x) = ln 4^x
(1-2x) ln(3) = x ln 4

multiply out using distributive)
ln 3 - 2x ln3 = x ln4 , bring x to one side

ln3 = xln4 + 2x ln3, factor out x
ln3 = x(ln4 + 2ln3)
x(ln4 + 2ln3) = ln3
x = ln3/(ln4 + 2ln3)

Answer 2

You should put parenthesis
3^(1-2x)=4^x
(1-2x)ln3 =xln4
x(2ln3+ln4)= ln3 so
x=ln3/(2ln3+ln4)

Answer 3

(1 - 2x) ln 3 = x ln 4
[(1 - 2x) ln 3] / x = ln 4
ln 3 / x - 2x ln 3 / x = ln 4
ln 3 / x - 2 ln 3 = ln 4
ln 3 / x = ln 4 + 2 ln 3
x = ln 3 / (ln 4 + 2 ln 3)