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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!

Blue
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Mevve
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2 Answers2

3

HINT

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

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

user
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  • Thank you for your answer! I wrote in the wrong equation. Sorry but it should be a • and not + between the natural logarithms. Sorry. – Mevve Sep 26 '18 at 10:14
  • @Mevve Ok I update that! – user Sep 26 '18 at 10:15
  • @Mevve The key point is always that $$\log a^b=b\log a$$ Are you aware about that? Try to prove that by the definition. It is a useful exercise. – user Sep 26 '18 at 10:17
  • Yes I know that, by using the definition I came up with my first answer witch is $x=e^{1/36}$ but that is wrong according to the answer sheet. – Mevve Sep 26 '18 at 10:32
  • From here we have $$(9\log x)\cdot (6\log x) =3 \iff (\log x)^2=\frac1{18} \iff \log x =\frac1{3\sqrt 2}$$ – user Sep 26 '18 at 10:34
  • Thank you, in my calculation I diveded by 2 again but I understand now that it is $(ln x)^2$ and not $(ln x^2)$. – Mevve Sep 26 '18 at 10:43
  • Yes of course we have $$\log x\cdot \log x=(\log x)^2=\log^2x$$ note that the last convention is usually adopted to indicate powers for log function. – user Sep 26 '18 at 11:33
1

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: $$\bbox[5px,border:2px solid red] {x = e^{\pm\sqrt{\frac{1}{18}}}}$$

Eff
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