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)=\frac52`

or

`\log_x9+\log_(x^2)3=\frac52`

• 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}`