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I need this equation to be solved for $x$. I know that I have to get rid of this natural logarithm in the exponent but I don't know how, nothing seems to work.

Is it possible to find $x$? Could you give me some hints? Is there something I'm missing out on?

$$203=412\,x^{1,265-0,0954\ln x}$$

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  • $\begingroup$ Hint : take $\ln$ of the equation & solve for $\ln$. $\endgroup$ Commented Nov 29, 2019 at 21:01
  • $\begingroup$ @DonaldSplutterwit I've exapanded your hint with the key point which justify that, i.e. the injectivity for log funtion. $\endgroup$ Commented Nov 29, 2019 at 21:09

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As suggested in the comments by Donald Splutterwit, noting also that $\log$ function is injective, we have that

$$203=412\,x^{1,265-0,0954\ln x} \iff \ln (203)=\ln \left(412\,x^{1,265-0,0954\ln x}\right)$$

then refer to the algebraic foundamental properties of logarithm

  • $\log (AB) = \log A+\log B$

  • $\log (A^B) = B\log A$

and solve the new expression obtained for $t=\ln x$ with the condition $x>0$.

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