Question-190961

Question Number 190961 by pascal889 last updated on 15/Apr/23

Answered by cortano12 last updated on 16/Apr/23

⇒log _(10) (3x^2 +8)=log _(10) (5x+10) ⇒3x^2 −5x−2=0 ⇒(3x+1)(x−2)=0 ⇒ { ((x=−(1/3))),((x=2)) :}

$$\:\Rightarrow\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{3x}^{\mathrm{2}} +\mathrm{8}\right)=\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{5x}+\mathrm{10}\right) \\ $$$$\:\Rightarrow\mathrm{3x}^{\mathrm{2}} −\mathrm{5x}−\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{3x}+\mathrm{1}\right)\left(\mathrm{x}−\mathrm{2}\right)=\mathrm{0}\: \\ $$$$\:\Rightarrow\begin{cases}{\mathrm{x}=−\frac{\mathrm{1}}{\mathrm{3}}}\\{\mathrm{x}=\mathrm{2}}\end{cases} \\ $$

Commented by pascal889 last updated on 15/Apr/23

$${please}\:{i}\:{dont}\:{really}\:{get}\:{ur}\:{workings} \\ $$$$ \\ $$

Answered by manxsol last updated on 15/Apr/23

log(3x^2 +8)=log10+log((x/2)+1) x>−2 log(3x^2 +8)=log(5x+10) 3x^2 +8=5x+10 3x^2 −5x−2=0 (3x+1)(x−2)=0 x=−(1/3) x =2

$${log}\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{8}\right)={log}\mathrm{10}+{log}\left(\frac{{x}}{\mathrm{2}}+\mathrm{1}\right) \\ $$$${x}>−\mathrm{2} \\ $$$${log}\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{8}\right)={log}\left(\mathrm{5}{x}+\mathrm{10}\right) \\ $$$$\mathrm{3}{x}^{\mathrm{2}} +\mathrm{8}=\mathrm{5}{x}+\mathrm{10} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{2}=\mathrm{0} \\ $$$$\left(\mathrm{3}{x}+\mathrm{1}\right)\left({x}−\mathrm{2}\right)=\mathrm{0} \\ $$$${x}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${x}\:=\mathrm{2} \\ $$

Answered by Rasheed.Sindhi last updated on 15/Apr/23

log_(10) (3x^2 +8)−log_(10) (((x+2)/2))=log_(10) 10 log_(10) ((3x^2 +8)/( ((x+2)/2) ))=log_(10) 10 ((3x^2 +8)/( ((x+2)/2) ))=10 3x^2 +8=10(((x+2)/2))=5x+10 3x^2 −5x−2=0 (x−2)(3x+1)=0 x=2 or x=−(1/3)

$$\mathrm{log}_{\mathrm{10}} \left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{8}\right)−\mathrm{log}_{\mathrm{10}} \left(\frac{{x}+\mathrm{2}}{\mathrm{2}}\right)=\mathrm{log}_{\mathrm{10}} \mathrm{10} \\ $$$$\mathrm{log}_{\mathrm{10}} \frac{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{8}}{\:\frac{{x}+\mathrm{2}}{\mathrm{2}}\:}=\mathrm{log}_{\mathrm{10}} \mathrm{10} \\ $$$$\frac{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{8}}{\:\frac{{x}+\mathrm{2}}{\mathrm{2}}\:}=\mathrm{10} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} +\mathrm{8}=\mathrm{10}\left(\frac{{x}+\mathrm{2}}{\mathrm{2}}\right)=\mathrm{5}{x}+\mathrm{10} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{2}=\mathrm{0} \\ $$$$\left({x}−\mathrm{2}\right)\left(\mathrm{3}{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=\mathrm{2}\:{or}\:{x}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Commented by pascal889 last updated on 15/Apr/23

$${thanks}\:{sir}\: \\ $$

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