Find the solution set of the equation

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

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

flatantsmu
2022-09-06
Answered

Find the solution set of the equation

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

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

You can still ask an expert for help

Emilia Boyle

Answered 2022-09-07
Author has **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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So $dv=\mathrm{ln}(4x)dx$ and $v=1/x$, but I don't know where to go from here.

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Here $\mathrm{ln}$ refers to the natural logarithm.

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How to simply this logarithmic equation?

I have

$f(L)={M}^{L-1}/(M+1{)}^{L}$

and

$L={\mathrm{log}}_{M}((K+B)/A)$

I am suppose to simply this to

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with

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I have

$f(L)={M}^{L-1}/(M+1{)}^{L}$

and

$L={\mathrm{log}}_{M}((K+B)/A)$

I am suppose to simply this to

$f=C(K+B{)}^{-b}$

with

$b={\displaystyle \frac{\mathrm{ln}(M+1)}{\mathrm{ln}(M)}}$

for the top I have simplified ${M}^{L-1}$ to $\frac{K+B}{AM}$, but I have no idea how to simplify the bottom part. Some help would be great

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How do I solve ${3}^{\mathrm{ln}2}\times {x}^{\mathrm{ln}x+\mathrm{ln}6+1}=\frac{3{e}^{2}}{4}$

I've been going at this question for 2 hours, my teacher wants us to solve for x without a graphing calculator.

$${3}^{\mathrm{ln}2}\times {x}^{\mathrm{ln}x+\mathrm{ln}6+1}=\frac{3{e}^{2}}{4}$$

I've been going at this question for 2 hours, my teacher wants us to solve for x without a graphing calculator.

$${3}^{\mathrm{ln}2}\times {x}^{\mathrm{ln}x+\mathrm{ln}6+1}=\frac{3{e}^{2}}{4}$$