Question

Solve for x if log 9 base x + log 3 base x^2 = 2.5​

Answer 1

Not sure if the equation is

$\log_9x+\log_3(x^2)=\dfrac52$

or

$\log_x9+\log_{x^2}3=\dfrac52$

• If it's the first one:

$9^{\log_9x+\log_3(x^2)}=9^{\log_9x}\cdot9^{\log_3(x^2)}$

$9^{\log_9x+\log_3(x^2)}=9^{\log_9x}\cdot(3^2)^{\log_3(x^2)}$

$9^{\log_9x+\log_3(x^2)}=9^{\log_9x}\cdot3^{2\log_3(x^2)}$

$9^{\log_9x+\log_3(x^2)}=9^{\log_9x}\cdot3^{\log_3(x^2)^2}$

$9^{\log_9x+\log_3(x^2)}=9^{\log_9x}\cdot3^{\log_3(x^4)}$

$9^{\log_9x+\log_3(x^2)}=x\cdot x^4$

$9^{\log_9x+\log_3(x^2)}=x^5$

On the other side of the equation, we'd get

$9^{5/2}=(3^2)^{5/2}=3^{2\cdot(5/2)}=3^5$

Then

$x^5=3^5\implies\boxed{x=3}$

• If it's the second one instead, you can use the same strategy as above:

$x^{\log_x9+\log_{x^2}3}=x^{\log_x9}\cdot x^{\log_{x^2}3}$

$x^{\log_x9+\log_{x^2}3}=x^{\log_x9}\cdot\left((x^2)^{1/2}\right)^{\log_{x^2}3}$

(Note that this step assume $x>0$)

$x^{\log_x9+\log_{x^2}3}=x^{\log_x9}\cdot(x^2)^{(1/2)\log_{x^2}3}$

$x^{\log_x9+\log_{x^2}3}=x^{\log_x9}\cdot(x^2)^{\log_{x^2}\sqrt3}$

$x^{\log_x9+\log_{x^2}3}=9\sqrt3$

Then we get

$9\sqrt3=x^{5/2}\implies x=(9\sqrt3)^{2/5}\implies\boxed{x=3}$