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I have an equation as follows:

$\ln(a)x + \ln(b) (1 - x) = \ln(c)$

I'd like to solve for $x$, but I think what I've come up with is wrong

$\ln\left(\frac{a}{b}\right)x + \ln(b) = \ln(c)$

$\ln\left(\frac{a}{b}\right)x = ln\left(\frac{c}{b}\right)$

$x = \frac{\ln\left(\frac{c}{b}\right)}{\ln\left(\frac{a}{b}\right)}$

Element118
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polyhedron
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1 Answers1

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$$x\ln a+(1-x)\ln b=\ln c$$ $$x\log_ba+1-x=\log_bc$$ $$x(\log_ba-1)=\log_bc-1$$ $$x=\frac{\log_bc-1}{\log_ba-1}=\frac{\log_b(\frac cb)}{\log_b(\frac ab)}$$

Aditya Agarwal
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  • You could even avoid changing base. Cheers – Claude Leibovici Jan 02 '16 at 05:42
  • Yea, but it was just to look good! Cheers! @ClaudeLeibovici. – Aditya Agarwal Jan 02 '16 at 05:43
  • @ClaudeLeibovici: Unfortunately, this answer is wrong because of the change of bases. For example try $(a,b) = (e,1)$. Then $x = \ln(c)$ but the expression in this answer is undefined. The morals of the story are: (1) Do not do more than you need to, otherwise you risk needing more than you want to; (2) Do not use identities before you check the requirements. =) – user21820 Jul 09 '16 at 06:57