Question-53205

Question Number 53205 by mondodotto@gmail.com last updated on 19/Jan/19

Commented by MJS last updated on 22/Jan/19

2 is a solution t_1 (x)=((25^x )/5^x )=(((25)/5))^x =5^x ; t_1 (2)=25 t_2 (x)=5^(x^2 −2) =(5^((x^2 )) /(25)); t_2 (2)=25 t_3 (x)=(√(50^x ))=((√(50)))^x =(5(√2))^x =5^x 2^(x/2) ; t_3 (x)=50

$$\mathrm{2}\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution} \\ $$$${t}_{\mathrm{1}} \left({x}\right)=\frac{\mathrm{25}^{{x}} }{\mathrm{5}^{{x}} }=\left(\frac{\mathrm{25}}{\mathrm{5}}\right)^{{x}} =\mathrm{5}^{{x}} ;\:{t}_{\mathrm{1}} \left(\mathrm{2}\right)=\mathrm{25} \\ $$$${t}_{\mathrm{2}} \left({x}\right)=\mathrm{5}^{{x}^{\mathrm{2}} −\mathrm{2}} =\frac{\mathrm{5}^{\left({x}^{\mathrm{2}} \right)} }{\mathrm{25}};\:{t}_{\mathrm{2}} \left(\mathrm{2}\right)=\mathrm{25} \\ $$$${t}_{\mathrm{3}} \left({x}\right)=\sqrt{\mathrm{50}^{{x}} }=\left(\sqrt{\mathrm{50}}\right)^{{x}} =\left(\mathrm{5}\sqrt{\mathrm{2}}\right)^{{x}} =\mathrm{5}^{{x}} \mathrm{2}^{\frac{{x}}{\mathrm{2}}} ;\:{t}_{\mathrm{3}} \left({x}\right)=\mathrm{50} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jan/19

((25^x )/5^x )+5^(x^2 −2) =(√(50^x )) (5^(2x) /5^x )+5^(x^2 −2) =(2×5^2 )^(x/2) 5^x +(5^x^2 /5^2 )=2^(x/2) ×5^x 5^x +5^(x^2 −2) −2^(x/2) ×5^x =0 5^x (1+5^(x^2 −2) −2^(x/2) )=0 if 5^x =0 =5^(−∞ ) x=−∞ when 1+5^(x^2 −2) −2^(x/2) =0 1+5^(x^2 −2) =2^(x/2) now taking the help of graph...

$$\frac{\mathrm{25}^{{x}} }{\mathrm{5}^{{x}} }+\mathrm{5}^{{x}^{\mathrm{2}} −\mathrm{2}} =\sqrt{\mathrm{50}^{{x}} } \\ $$$$\frac{\mathrm{5}^{\mathrm{2}{x}} }{\mathrm{5}^{{x}} }+\mathrm{5}^{{x}^{\mathrm{2}} −\mathrm{2}} =\left(\mathrm{2}×\mathrm{5}^{\mathrm{2}} \right)^{\frac{{x}}{\mathrm{2}}} \\ $$$$\mathrm{5}^{{x}} +\frac{\mathrm{5}^{{x}^{\mathrm{2}} } }{\mathrm{5}^{\mathrm{2}} }=\mathrm{2}^{\frac{{x}}{\mathrm{2}}} ×\mathrm{5}^{{x}} \\ $$$$\mathrm{5}^{{x}} +\mathrm{5}^{{x}^{\mathrm{2}} −\mathrm{2}} −\mathrm{2}^{\frac{{x}}{\mathrm{2}}} ×\mathrm{5}^{{x}} =\mathrm{0} \\ $$$$\mathrm{5}^{{x}} \left(\mathrm{1}+\mathrm{5}^{{x}^{\mathrm{2}} −\mathrm{2}} −\mathrm{2}^{\frac{{x}}{\mathrm{2}}} \right)=\mathrm{0} \\ $$$${if} \\ $$$$\mathrm{5}^{{x}} =\mathrm{0}\:=\mathrm{5}^{−\infty\:} \:\:{x}=−\infty \\ $$$${when} \\ $$$$\mathrm{1}+\mathrm{5}^{{x}^{\mathrm{2}} −\mathrm{2}} −\mathrm{2}^{\frac{{x}}{\mathrm{2}}} =\mathrm{0} \\ $$$$\mathrm{1}+\mathrm{5}^{{x}^{\mathrm{2}} −\mathrm{2}} =\mathrm{2}^{\frac{{x}}{\mathrm{2}}} \\ $$$${now}\:{taking}\:{the}\:{help}\:{of}\:{graph}… \\ $$$$ \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 19/Jan/19

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