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Question Number 104199 by hardylanes last updated on 20/Jul/20

solve for real values of x the equation 4(3^(2x+1) )+17(3^x )=7. if m and n are positive real numbers other than 1, show that the log_n m+log_(1/m) n=0

$${solve}\:{for}\:{real}\:{values}\:{of}\:{x}\:{the}\:{equation} \\ $$$$\mathrm{4}\left(\mathrm{3}^{\mathrm{2}{x}+\mathrm{1}} \right)+\mathrm{17}\left(\mathrm{3}^{{x}} \right)=\mathrm{7}. \\ $$$${if}\:{m}\:{and}\:{n}\:{are}\:{positive}\:{real}\:{numbers}\:{other} \\ $$$${than}\:\mathrm{1},\:{show}\:{that}\:{the}\:\mathrm{log}_{{n}} {m}+\mathrm{log}_{\frac{\mathrm{1}}{{m}}} {n}=\mathrm{0} \\ $$

Answered by bemath last updated on 20/Jul/20

12.3^x +17.3^x −7=0 3^x = ((−17+(√(289+48.7)))/(24)) 3^x = ((−17+25)/(24)) = (1/3) ∴ x = −1

$$\mathrm{12}.\mathrm{3}^{{x}} +\mathrm{17}.\mathrm{3}^{{x}} −\mathrm{7}=\mathrm{0} \\ $$$$\mathrm{3}^{{x}} =\:\frac{−\mathrm{17}+\sqrt{\mathrm{289}+\mathrm{48}.\mathrm{7}}}{\mathrm{24}} \\ $$$$\mathrm{3}^{{x}} \:=\:\frac{−\mathrm{17}+\mathrm{25}}{\mathrm{24}}\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\therefore\:{x}\:=\:−\mathrm{1} \\ $$

Answered by OlafThorendsen last updated on 20/Jul/20

4(3^(2x+1) )+17(3^x ) = 7 12(3^x )^2 +17(3^x )−7 = 0 u = 3^x 12u^2 +17u−7 = 0 Δ = 17^2 −4×12×(−7) = 625 = 25^2 u = ((−17±25)/(24)) u = 3^x ⇒ u>0 then u = ((−17+25)/(24)) = (1/3) u = 3^x = (1/3) ⇒ x = −1 I think your second question is false (it′s true only for m = n or m = (1/n)) but : log_n m+log_(1/n) m = ((lnm)/(lnn))+((lnm)/(ln(1/n))) = ((lnm)/(lnn))−((lnm)/(lnn)) = 0

$$\mathrm{4}\left(\mathrm{3}^{\mathrm{2}{x}+\mathrm{1}} \right)+\mathrm{17}\left(\mathrm{3}^{{x}} \right)\:=\:\mathrm{7} \\ $$$$\mathrm{12}\left(\mathrm{3}^{{x}} \right)^{\mathrm{2}} +\mathrm{17}\left(\mathrm{3}^{{x}} \right)−\mathrm{7}\:=\:\mathrm{0} \\ $$$${u}\:=\:\mathrm{3}^{{x}} \\ $$$$\mathrm{12}{u}^{\mathrm{2}} +\mathrm{17}{u}−\mathrm{7}\:=\:\mathrm{0} \\ $$$$\Delta\:=\:\mathrm{17}^{\mathrm{2}} −\mathrm{4}×\mathrm{12}×\left(−\mathrm{7}\right)\:=\:\mathrm{625}\:=\:\mathrm{25}^{\mathrm{2}} \\ $$$${u}\:=\:\frac{−\mathrm{17}\pm\mathrm{25}}{\mathrm{24}} \\ $$$${u}\:=\:\mathrm{3}^{{x}} \:\Rightarrow\:{u}>\mathrm{0} \\ $$$$\mathrm{then}\:{u}\:=\:\frac{−\mathrm{17}+\mathrm{25}}{\mathrm{24}}\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${u}\:=\:\mathrm{3}^{{x}} \:=\:\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\:{x}\:=\:−\mathrm{1} \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{your}\:\mathrm{second} \\ $$$$\mathrm{question}\:\mathrm{is}\:\mathrm{false} \\ $$$$\left(\mathrm{it}'\mathrm{s}\:\mathrm{true}\:\mathrm{only}\:\mathrm{for}\:{m}\:=\:{n}\right. \\ $$$$\left.\mathrm{or}\:{m}\:=\:\frac{\mathrm{1}}{{n}}\right)\:\mathrm{but}\:: \\ $$$$\mathrm{log}_{{n}} {m}+\mathrm{log}_{\frac{\mathrm{1}}{{n}}} {m}\:= \\ $$$$\frac{\mathrm{ln}{m}}{\mathrm{ln}{n}}+\frac{\mathrm{ln}{m}}{\mathrm{ln}\frac{\mathrm{1}}{{n}}}\:=\: \\ $$$$\frac{\mathrm{ln}{m}}{\mathrm{ln}{n}}−\frac{\mathrm{ln}{m}}{\mathrm{ln}{n}}\:=\:\mathrm{0} \\ $$

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