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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

 Apr 7, 2016
edited by MWizard2k04  Apr 7, 2016
edited by MWizard2k04  Apr 7, 2016
edited by MWizard2k04  Apr 7, 2016
 #1
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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

 Apr 7, 2016

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