flatantsmu

2022-09-06

Find the solution set of the equation

$${\mathrm{log}}_{x}(3x-2)+{\mathrm{log}}_{{x}^{2}}(\frac{1}{3x-2})=1$$

$${\mathrm{log}}_{x}(3x-2)+{\mathrm{log}}_{{x}^{2}}(\frac{1}{3x-2})=1$$

Emilia Boyle

Beginner2022-09-07Added 10 answers

Given

$${\mathrm{log}}_{x}(3x-2)+{\mathrm{log}}_{{x}^{2}}(\frac{1}{3x-2})=1\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{log}}_{x}(3x-2)+\frac{1}{2}({\mathrm{log}}_{x}1-{\mathrm{log}}_{x}(3x-2))=1\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{log}}_{x}(3x-2)+\frac{1}{2}(0-{\mathrm{log}}_{x}(3x-2))=1\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{log}}_{x}(3x-2)-\frac{1}{2}{\mathrm{log}}_{x}(3x-2)=1\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}{\mathrm{log}}_{x}(3x-2)=1\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{log}}_{x}(3x-2)=2\phantom{\rule{0ex}{0ex}}\Rightarrow (3x-2)={x}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-3x+2=0\phantom{\rule{0ex}{0ex}}\Rightarrow x(x-2)-1(x-2)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (x-1)(x-2)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x-1=0\text{or}x-2=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=1\text{or}x=2$$

but $$x\ne 1$$ as x is in base and nase $$\ne 1$$ x=2 is only solution.

$${\mathrm{log}}_{x}(3x-2)+{\mathrm{log}}_{{x}^{2}}(\frac{1}{3x-2})=1\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{log}}_{x}(3x-2)+\frac{1}{2}({\mathrm{log}}_{x}1-{\mathrm{log}}_{x}(3x-2))=1\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{log}}_{x}(3x-2)+\frac{1}{2}(0-{\mathrm{log}}_{x}(3x-2))=1\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{log}}_{x}(3x-2)-\frac{1}{2}{\mathrm{log}}_{x}(3x-2)=1\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}{\mathrm{log}}_{x}(3x-2)=1\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{log}}_{x}(3x-2)=2\phantom{\rule{0ex}{0ex}}\Rightarrow (3x-2)={x}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-3x+2=0\phantom{\rule{0ex}{0ex}}\Rightarrow x(x-2)-1(x-2)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (x-1)(x-2)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x-1=0\text{or}x-2=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=1\text{or}x=2$$

but $$x\ne 1$$ as x is in base and nase $$\ne 1$$ x=2 is only solution.

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