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Question Number 5764 by sanusihammed last updated on 26/May/16

2^x = 4x Solution 2^x = 4x This can be re write as (1+1)^x = 4x Using combination to epand from the identity. (1+x)^n = 1+nx+((n(n−1))/(2!))x^2 +((n(n−1)(n−2))/(3!))x^3 +....+nCrx^(r ) Therefore. (1+1)^x = 4x 1+x+((x(x−1))/(2!))(1^2 )+((x(x−1)(x−2))/(3!))(1^3 )+......+1 = 4x ignore the continuity (what rule is ..... ignore the +...+) is it linear approximation 1+x+((x^2 −x)/(2×1))+(((x^2 −x)(x−2))/(3×2×1))+1 = 4x 1+x+((x^2 −x)/2)+((x^3 −2x^2 −x^2 +2x)/6)+1 = 4x 1+x+((x^2 −x)/2)+((x^3 −3x^2 +2x)/6)+1 = 4x Multiply through by 6 6+6x+3(x^2 −x)+x^3 −3x^2 +2x+6 = 24x 12+6x+3x^2 −3x+x^3 −3x^2 +2x+6 = 24x 12+5x+x^3 = 24x 12+5x+x^3 −24x = 0 x^3 −19x+12 = 0 Factorize x^3 −4x^2 +4x^2 −16x−3x+12 = 0 (x^3 −4x^2 )+(4x^2 −16x)−(3x+12) = 0 x^2 (x−4)+4x(x−4)−3(x−4) = 0 Factor out (x−4) (x−4)(x^2 +4x−3) = 0 x−4 = 0 or x^( 2) +4x−3 = 0 x = 4 or x = 0.6458 or x = −4.6458 The only real solution is x = 4 Therefore x = 4 DONE! Please confirm the solution. is it correct or please corect it or show me alternative. This is my trial. Thanks.

$$\mathrm{2}^{{x}} \:=\:\mathrm{4}{x} \\ $$$${Solution} \\ $$$$\mathrm{2}^{{x}} \:=\:\mathrm{4}{x} \\ $$$${This}\:{can}\:{be}\:{re}\:{write}\:{as} \\ $$$$\left(\mathrm{1}+\mathrm{1}\right)^{{x}} \:=\:\mathrm{4}{x} \\ $$$${Using}\:{combination}\:{to}\:{epand} \\ $$$${from}\:{the}\:{identity}. \\ $$$$\left(\mathrm{1}+{x}\right)^{{n}} \:=\:\mathrm{1}+{nx}+\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}!}{x}^{\mathrm{2}} +\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{3}!}{x}^{\mathrm{3}} +....+{nCrx}^{{r}\:} \: \\ $$$${Therefore}. \\ $$$$\left(\mathrm{1}+\mathrm{1}\right)^{{x}} \:=\:\mathrm{4}{x} \\ $$$$\mathrm{1}+{x}+\frac{{x}\left({x}−\mathrm{1}\right)}{\mathrm{2}!}\left(\mathrm{1}^{\mathrm{2}} \right)+\frac{{x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)}{\mathrm{3}!}\left(\mathrm{1}^{\mathrm{3}} \right)+......+\mathrm{1}\:=\:\mathrm{4}{x} \\ $$$${ignore}\:{the}\:{continuity}\:\left({what}\:{rule}\:{is}\:.....\:{ignore}\:{the}\:+...+\right)\:{is}\:{it} \\ $$$${linear}\:{approximation} \\ $$$$\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} −{x}}{\mathrm{2}×\mathrm{1}}+\frac{\left({x}^{\mathrm{2}} −{x}\right)\left({x}−\mathrm{2}\right)}{\mathrm{3}×\mathrm{2}×\mathrm{1}}+\mathrm{1}\:=\:\mathrm{4}{x} \\ $$$$\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} −{x}}{\mathrm{2}}+\frac{{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} −{x}^{\mathrm{2}} +\mathrm{2}{x}}{\mathrm{6}}+\mathrm{1}\:=\:\mathrm{4}{x} \\ $$$$\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} −{x}}{\mathrm{2}}+\frac{{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}}{\mathrm{6}}+\mathrm{1}\:=\:\mathrm{4}{x} \\ $$$${Multiply}\:{through}\:{by}\:\mathrm{6}\: \\ $$$$\mathrm{6}+\mathrm{6}{x}+\mathrm{3}\left({x}^{\mathrm{2}} −{x}\right)+{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{6}\:=\:\mathrm{24}{x} \\ $$$$\mathrm{12}+\mathrm{6}{x}+\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3}{x}+{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{6}\:=\:\mathrm{24}{x} \\ $$$$\mathrm{12}+\mathrm{5}{x}+{x}^{\mathrm{3}} \:=\:\mathrm{24}{x} \\ $$$$\mathrm{12}+\mathrm{5}{x}+{x}^{\mathrm{3}} −\mathrm{24}{x}\:=\:\mathrm{0} \\ $$$${x}^{\mathrm{3}} −\mathrm{19}{x}+\mathrm{12}\:=\:\mathrm{0} \\ $$$${Factorize} \\ $$$${x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{16}{x}−\mathrm{3}{x}+\mathrm{12}\:=\:\mathrm{0} \\ $$$$\left({x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} \right)+\left(\mathrm{4}{x}^{\mathrm{2}} −\mathrm{16}{x}\right)−\left(\mathrm{3}{x}+\mathrm{12}\right)\:=\:\mathrm{0} \\ $$$${x}^{\mathrm{2}} \left({x}−\mathrm{4}\right)+\mathrm{4}{x}\left({x}−\mathrm{4}\right)−\mathrm{3}\left({x}−\mathrm{4}\right)\:=\:\mathrm{0} \\ $$$${Factor}\:{out}\:\left({x}−\mathrm{4}\right) \\ $$$$\left({x}−\mathrm{4}\right)\left({x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{3}\right)\:=\:\mathrm{0} \\ $$$${x}−\mathrm{4}\:=\:\mathrm{0}\:{or}\:{x}^{\:\mathrm{2}} +\mathrm{4}{x}−\mathrm{3}\:=\:\mathrm{0} \\ $$$${x}\:=\:\mathrm{4}\:{or}\:{x}\:=\:\mathrm{0}.\mathrm{6458}\:{or}\:{x}\:=\:−\mathrm{4}.\mathrm{6458} \\ $$$${The}\:{only}\:{real}\:{solution}\:{is}\:{x}\:=\:\mathrm{4} \\ $$$${Therefore} \\ $$$${x}\:=\:\mathrm{4} \\ $$$$ \\ $$$${DONE}! \\ $$$$ \\ $$$${Please}\:{confirm}\:{the}\:{solution}.\:{is}\:{it}\:{correct}\:{or}\:{please}\:{corect}\:{it}\:{or} \\ $$$${show}\:{me}\:{alternative}. \\ $$$${This}\:{is}\:{my}\:{trial}. \\ $$$${Thanks}. \\ $$$$ \\ $$$$ \\ $$

Commented by prakash jain last updated on 26/May/16

(1+x)^n =Σ_(i=0) ^n ^n C_i x^i only if n∈Z

$$\left(\mathrm{1}+{x}\right)^{{n}} =\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}\:^{{n}} {C}_{{i}} {x}^{{i}} \:\mathrm{only}\:\mathrm{if}\:{n}\in\mathbb{Z} \\ $$

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