$$3^{2-3x} = 4^{2x+1}$$
Use common log and two decimal place approx.

Question

$$3^{2-3x} = 4^{2x+1}$$   
Use common log and two decimal place approx.

mani_jha · Answer

(2-3x)log 3=(2x+1)log 4
Plug in the values of log 3 and log 4. Then it's just a linear equation to solve.

anonymous · Answer

I get:  $$-x= \log 3^{-2} x 4  \div 3^{3}\div 4^{2} $$

mani_jha · Answer

$$2\log_{}3-3x \log_{}3=2x \log_{}4+\log_{}4    $$
$$\log 3^{2}-\log4=2xlog 4+3xlog3$$
$$\log_{} (9/4)=x(\log 16+\log 27)$$
x=$$(\log(9/4)/\log(27/16)$$
 How did you get that? o.o

anonymous · Answer

Wasn't sure what to do with that "-x" when I pulled it out early on in the equation.  Then  I left in an extra x (before the 4.)   So I'm confused about pulling out the x early on... do I keep that negative with the 3 log 3  - 2 log 4?

mani_jha · Answer

yes. don't keep it with  x