Question
Solve the following equations for $x$: $3^{2\text{x}+4}+1=2\times3^{\text{x}+2}$
✓
Answer
We have,$3^{2\text{x}+4}+1=2\times3^{\text{x}+2}$
$(3^{\text{x}+2})^2+1=2\times3^{\text{x}+2}$
$\text{Let}\ 3^{\text{x}+2}=\text{y}$
$\text{y}^2+1=2\text{y}$
$\text{y}^2=2\text{y}+1=0$
$\text{y}^2-\text{y}-\text{y}+1=0$
$\text{y}(\text{y}-1)-1(\text{y}-1)=0$
$(\text{y}-1)(\text{y}-1)=0$
$\text{y}=1$
$(3^{\text{x}+2})^2+1=2\times3^{\text{x}+2}$
$\text{Let}\ 3^{\text{x}+2}=\text{y}$
$\text{y}^2+1=2\text{y}$
$\text{y}^2=2\text{y}+1=0$
$\text{y}^2-\text{y}-\text{y}+1=0$
$\text{y}(\text{y}-1)-1(\text{y}-1)=0$
$(\text{y}-1)(\text{y}-1)=0$
$\text{y}=1$
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