So I've been stuck with this equation - $\log_{\frac{x}{2}}(x^2) - 14\log_{16x}(x^3) + 40\log_{4x}(\sqrt{x}) = 0$. I was thinking of using identities such as $\log_{a}(x) = \frac{\log_{b}(x)}{\log_{b}(a)}$ but that didn't simplify much. Do you have an idea on how to proceed?
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$$\log_{\frac{x}{2}}(x^2) - 14\log_{16x}(x^3) + 40\log_{4x}(\sqrt{x}) = 0$$ $$\frac{\ln(x^2)}{\ln(x/2)}-14\frac{\ln(x^3)}{\ln(16x)}+40\frac{\ln \sqrt x}{\ln(4x)}=0$$ $$\frac{2\ln x}{\ln x-\ln 2}-14\frac{3\ln x}{\ln 16+\ln x}+40\frac{\frac{1}{2}\ln x}{\ln 4+\ln x}=0$$ Now set $\ln x=u$ and solve the algebraic equation.
Remember that $x>0;\;x\ne 2;\;x\ne \frac{1}{16};\;x\ne\frac{1}{4}$ otherwise the bases of the logarithm in the given equation become $1$ and this is not allowed
Raffaele
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Thank you very much. I got to the third line as well but forgot about using a substitution. – Thomas Petit Dec 09 '20 at 21:16
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@ThomasPetit You must add your work in the question! It's very important in this community, because otherwise people can think that you just want them to do your homework – Raffaele Dec 09 '20 at 21:24
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Apologies, I will keep that in mind and present my work next time – Thomas Petit Dec 09 '20 at 21:49