+426 3^(x+1)=4^(x-1) | *4
4*3^(x+1)=4^x
What's the meaning of this? (Extracted from the very first question)
http://web2.0calc.com/questions/3-x-1-4-x-1-solve-for-x
+118687 This was the question
3^(x+1)=4^(x-1)
This is the beginning of Admin's answer
3^(x+1)=4^(x-1) | *4
What Admin meant is that you have to multiply both sides by 4.
\(4* 3^{(x+1)}=4^{(x-1)} *4\\ 4* 3^{(x+1)}=4^{x-1} *4^1\\ 4* 3^{(x+1)}=4^{x-1+1} \\ 4* 3^{(x+1)}=4^{x} \\ 4* 3^{(x+1)}=4^{x} \\\)
Then he took the log of of both sides
\(ln(4* 3^{(x+1)})=ln(4^{x}) \\ ln(4)+ln( 3^{(x+1)})=ln(4^{x}) \\ ln(4)+(x+1)ln( 3)=xln(4) \\ ln4+xln3+ln3=xln4 \\ ln4+ln3=xln4-xln3 \\ ln4+ln3=x(ln4-ln3) \\ ln12=xln(4/3)\\ \frac{ln12}{ln(4/3)}=x\\ x=\frac{ln12}{ln(4/3)}\)
This is the same as admins answer of -ln12/(ln3-ln4) = 8.6376
