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Question Number 142755 by ERA last updated on 05/Jun/21

log_3 x^3 +log_2 x^2 =((2lg6)/(lg2))+1 find x

$$\mathrm{log}_{\mathrm{3}} \mathrm{x}^{\mathrm{3}} +\mathrm{log}_{\mathrm{2}} \mathrm{x}^{\mathrm{2}} =\frac{\mathrm{2lg6}}{\mathrm{lg2}}+\mathrm{1}\:\:\mathrm{find}\:\:\mathrm{x} \\ $$

Commented by iloveisrael last updated on 05/Jun/21

what different log and lg ?

$${what}\:{different}\:\mathrm{log}\:\:{and}\:\mathrm{lg}\:? \\ $$

Commented by qaz last updated on 05/Jun/21

lgx=log_(10) x

$$\mathrm{lgx}=\mathrm{log}_{\mathrm{10}} \mathrm{x} \\ $$

Answered by iloveisrael last updated on 05/Jun/21

⇒log _3 (x^3 )+log _2 (x^2 )=2log _2 (3×2)+log _3 (3) ⇒3log _3 (x)+2log _2 (x)=2log _2 (2)+3log _3 (3) ⇒3log_3 ((x/3))=2log_2 ((2/x)) ⇒3log_3 ((x/3))=−2log_2 ((x/2)) ⇒((3ln x−3ln 3)/(ln 3)) = ((−2ln x+2ln 2)/(ln 2)) ⇒3ln 2.ln x−3ln 2.ln 3=−2ln 3.ln x+2ln 2.ln 3 ⇒2ln x.ln 8=2ln 2.ln 8 ⇒x = 2

$$\Rightarrow\mathrm{log}\:_{\mathrm{3}} \left({x}^{\mathrm{3}} \right)+\mathrm{log}\:_{\mathrm{2}} \left({x}^{\mathrm{2}} \right)=\mathrm{2log}\:_{\mathrm{2}} \left(\mathrm{3}×\mathrm{2}\right)+\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{3}\right) \\ $$$$\Rightarrow\mathrm{3log}\:_{\mathrm{3}} \left({x}\right)+\mathrm{2log}\:_{\mathrm{2}} \left({x}\right)=\mathrm{2log}\:_{\mathrm{2}} \left(\mathrm{2}\right)+\mathrm{3log}\:_{\mathrm{3}} \left(\mathrm{3}\right) \\ $$$$\Rightarrow\mathrm{3log}_{\mathrm{3}} \:\left(\frac{{x}}{\mathrm{3}}\right)=\mathrm{2log}_{\mathrm{2}} \:\left(\frac{\mathrm{2}}{{x}}\right) \\ $$$$\Rightarrow\mathrm{3log}_{\mathrm{3}} \:\left(\frac{{x}}{\mathrm{3}}\right)=−\mathrm{2log}_{\mathrm{2}} \:\left(\frac{{x}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\frac{\mathrm{3ln}\:{x}−\mathrm{3ln}\:\mathrm{3}}{\mathrm{ln}\:\mathrm{3}}\:=\:\frac{−\mathrm{2ln}\:{x}+\mathrm{2ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{2}} \\ $$$$\Rightarrow\mathrm{3ln}\:\mathrm{2}.\mathrm{ln}\:{x}−\mathrm{3ln}\:\mathrm{2}.\mathrm{ln}\:\mathrm{3}=−\mathrm{2ln}\:\mathrm{3}.\mathrm{ln}\:{x}+\mathrm{2ln}\:\mathrm{2}.\mathrm{ln}\:\mathrm{3} \\ $$$$\Rightarrow\mathrm{2ln}\:{x}.\cancel{\mathrm{ln}\:\mathrm{8}}=\mathrm{2ln}\:\mathrm{2}.\cancel{\mathrm{ln}\:\mathrm{8}} \\ $$$$\Rightarrow{x}\:=\:\mathrm{2}\: \\ $$

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