Not sure if the equation is
[tex]\log_9x+\log_3(x^2)=\dfrac52[/tex]
or
[tex]\log_x9+\log_{x^2}3=\dfrac52[/tex]
[tex]9^{\log_9x+\log_3(x^2)}=9^{\log_9x}\cdot9^{\log_3(x^2)}[/tex]
[tex]9^{\log_9x+\log_3(x^2)}=9^{\log_9x}\cdot(3^2)^{\log_3(x^2)}[/tex]
[tex]9^{\log_9x+\log_3(x^2)}=9^{\log_9x}\cdot3^{2\log_3(x^2)}[/tex]
[tex]9^{\log_9x+\log_3(x^2)}=9^{\log_9x}\cdot3^{\log_3(x^2)^2}[/tex]
[tex]9^{\log_9x+\log_3(x^2)}=9^{\log_9x}\cdot3^{\log_3(x^4)}[/tex]
[tex]9^{\log_9x+\log_3(x^2)}=x\cdot x^4[/tex]
[tex]9^{\log_9x+\log_3(x^2)}=x^5[/tex]
On the other side of the equation, we'd get
[tex]9^{5/2}=(3^2)^{5/2}=3^{2\cdot(5/2)}=3^5[/tex]
Then
[tex]x^5=3^5\implies\boxed{x=3}[/tex]
- If it's the second one instead, you can use the same strategy as above:
[tex]x^{\log_x9+\log_{x^2}3}=x^{\log_x9}\cdot x^{\log_{x^2}3}[/tex]
[tex]x^{\log_x9+\log_{x^2}3}=x^{\log_x9}\cdot\left((x^2)^{1/2}\right)^{\log_{x^2}3}[/tex]
(Note that this step assume [tex]x>0[/tex])
[tex]x^{\log_x9+\log_{x^2}3}=x^{\log_x9}\cdot(x^2)^{(1/2)\log_{x^2}3}[/tex]
[tex]x^{\log_x9+\log_{x^2}3}=x^{\log_x9}\cdot(x^2)^{\log_{x^2}\sqrt3}[/tex]
[tex]x^{\log_x9+\log_{x^2}3}=9\sqrt3[/tex]
Then we get
[tex]9\sqrt3=x^{5/2}\implies x=(9\sqrt3)^{2/5}\implies\boxed{x=3}[/tex]
