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The equation is $(3/2)\ln x= -2$. I am not sure how to work this one. If anyone could show all the steps that would be a great help. I tried working it out and got down to $x^{3/2}=e^{-2}$ that is probably wrong but if by some chance it is right what do you do after that?

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  • $\begingroup$ Keep the $\ln x$ alone. Use the fact that $\exp(\ln x)=x$ $\endgroup$ – npisinp Apr 9 '14 at 22:37
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That $x^{3/2}=e^{-2}$ is correct. If you know $x^{3/2}=\text{something}$, then you can say $$ \Big(x^{3/2}\Big)^{2/3} = \text{something}^{2/3} $$ and then $$ x^{(3/2)\cdot(2/3)} = \text{something}^{2/3}. $$

Then remember how to simplify $\dfrac 3 2 \cdot \dfrac 2 3$.

Then you need to simplify $\displaystyle\Big(e^{-2}\Big)^{2/3}$. That is similar to something done above.

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  • $\begingroup$ Thank you so much! This clarifies everything! $\endgroup$ – Ila Isabelle Apr 9 '14 at 23:02
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$$ \frac{3}{2} \ln x = -2 \iff \ln x = \frac{ -4}{3} \iff e^{-\frac{4}{3} }= x$$

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  • $\begingroup$ Okay just a question, why would you not raise the x to (3/2)? My teacher said that is what we would do because it is a logarithm property. $\endgroup$ – Ila Isabelle Apr 9 '14 at 22:43
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$$\dfrac{3}{2}\ln x=-2\\ \implies \ln x=\dfrac{-4}{3}\\ \implies \exp(\ln x)=\boxed{x=\exp\left(\dfrac{-4}{3}\right)}$$

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  • $\begingroup$ "$=-2$" and not "$=-1$". $\endgroup$ – ZHN Apr 9 '14 at 22:44
  • $\begingroup$ So do you multiply the (3/2) to the other side being the -2? and this is probably a stupid question but what does exp stand for? $\endgroup$ – Ila Isabelle Apr 9 '14 at 22:54
  • $\begingroup$ @IlaIsabelle $$\exp(x)=e^x$$ $\endgroup$ – user122283 Apr 9 '14 at 22:57
  • $\begingroup$ Oh thank you!!! $\endgroup$ – Ila Isabelle Apr 9 '14 at 22:59
  • $\begingroup$ @IlaIsabelle Sure! $\endgroup$ – user122283 Apr 9 '14 at 22:59

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