Solving $ \ln(x^9) \cdot \ln(x^6)=3 $

Question

I'm completely stuck on this equation, it feels like I have tried everything. Any tips would be appreciated.

$$ \ln(x^9) \cdot \ln(x^6)=3 $$

I have tried all kinds of possible solutions.

The farthest I have got (I think) is by rewriting the equation to:

$$ \frac{\log_x x^9}{\log_x e} \cdot \frac{\log_x x^6}{\log_x e} = \log_x x^3 $$ $$ \frac{9}{\log_x e} \cdot \frac{6}{\log_x e} = \log_x x^3 $$

I don't even know if I'm on the right track here. I'm in desperate mode at the moment.

Any help will be appreciated!

Answer 1

HINT

Recall that $$\log a^b=b\log a$$ then

$$\ln(x^9)\cdot \ln(x^6)=3\iff (9\log x)\cdot (6\log x) =3$$

Answer 2

For any logarithm we have that $\log (a^b) = b\cdot \log(a)$. Apply this, so that $$3=\ln(x^9)\cdot \ln(x^6) = \big(9\cdot\ln(x)\big)\cdot \big(6\cdot \ln(x)\big) $$

Divide by $9\cdot 6$ (to move all the constants over to one side and only logarithms on the other), and we get $$\frac{3}{9\cdot 6} = \frac{1}{18} = \ln(x)\cdot\ln(x) = \big(\ln(x)\big)^2 $$

Then applying square root we get $$\ln(x) = \pm \sqrt{\frac{1}{18}}$$

Applying the exponential function to the equation (the inverse operation of $\ln$), we get the two solutions: $$\bbox[5px,border:2px solid red] {x = e^{\pm\sqrt{\frac{1}{18}}}}$$