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Question Number 126619 by O Predador last updated on 22/Dec/20

     5^x  + 7^x  = (9.8)^x       x = ?

$$\: \\ $$$$\:\:\mathrm{5}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{7}^{\boldsymbol{\mathrm{x}}} \:=\:\left(\mathrm{9}.\mathrm{8}\right)^{\boldsymbol{\mathrm{x}}} \\ $$$$\: \\ $$$$\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$$$\: \\ $$

Answered by Dwaipayan Shikari last updated on 22/Dec/20

((5/(9.8)))^x +((7/(9.8)))^x =1  ⇒(((5/7))^x a)+a=1        a=((7/(9.8)))^x =(((10)/(14)))^x =((5/7))^x   ⇒a^2 +a−1=0⇒a=((−1±(√5))/2)=(((√5)−1)/2)  ((5/7))^x =(((√5)−1)/2)⇒x=((log((((√5)−1)/2)))/(log((5/7))))∼1.4301679980845866709...

$$\left(\frac{\mathrm{5}}{\mathrm{9}.\mathrm{8}}\right)^{{x}} +\left(\frac{\mathrm{7}}{\mathrm{9}.\mathrm{8}}\right)^{{x}} =\mathrm{1} \\ $$$$\Rightarrow\left(\left(\frac{\mathrm{5}}{\mathrm{7}}\right)^{{x}} {a}\right)+{a}=\mathrm{1}\:\:\:\:\:\:\:\:{a}=\left(\frac{\mathrm{7}}{\mathrm{9}.\mathrm{8}}\right)^{{x}} =\left(\frac{\mathrm{10}}{\mathrm{14}}\right)^{{x}} =\left(\frac{\mathrm{5}}{\mathrm{7}}\right)^{{x}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{a}−\mathrm{1}=\mathrm{0}\Rightarrow{a}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{5}}{\mathrm{7}}\right)^{{x}} =\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\Rightarrow{x}=\frac{{log}\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right)}{{log}\left(\frac{\mathrm{5}}{\mathrm{7}}\right)}\sim\mathrm{1}.\mathrm{4301679980845866709}... \\ $$

Commented by O Predador last updated on 22/Dec/20

Wonderful!

$${Wonderful}! \\ $$

Answered by Olaf last updated on 22/Dec/20

5^x +7^x  = 9.8^x  = (((7×14)/(5×2)))^x   5^x +7^x  = 9.8^x  = 7^x ((7/5))^x   ((5/7))^x +1 = 9.8^x  = ((7/5))^x   Let X = ((5/7))^x   X+1 = (1/X)  X^2 +X−1 = 0  X = ((−1±(√5))/2)  ((−1−(√5))/2) impossible because X>0  X = ((−1+(√5))/2) = ((5/7))^x   xln(5/7) = ln((((√5)−1)/2))  x = ((ln((((√5)−1)/2)))/(ln(5/7)))

$$\mathrm{5}^{{x}} +\mathrm{7}^{{x}} \:=\:\mathrm{9}.\mathrm{8}^{{x}} \:=\:\left(\frac{\mathrm{7}×\mathrm{14}}{\mathrm{5}×\mathrm{2}}\right)^{{x}} \\ $$$$\mathrm{5}^{{x}} +\mathrm{7}^{{x}} \:=\:\mathrm{9}.\mathrm{8}^{{x}} \:=\:\mathrm{7}^{{x}} \left(\frac{\mathrm{7}}{\mathrm{5}}\right)^{{x}} \\ $$$$\left(\frac{\mathrm{5}}{\mathrm{7}}\right)^{{x}} +\mathrm{1}\:=\:\mathrm{9}.\mathrm{8}^{{x}} \:=\:\left(\frac{\mathrm{7}}{\mathrm{5}}\right)^{{x}} \\ $$$$\mathrm{Let}\:\mathrm{X}\:=\:\left(\frac{\mathrm{5}}{\mathrm{7}}\right)^{{x}} \\ $$$$\mathrm{X}+\mathrm{1}\:=\:\frac{\mathrm{1}}{\mathrm{X}} \\ $$$$\mathrm{X}^{\mathrm{2}} +\mathrm{X}−\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{X}\:=\:\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\:\mathrm{impossible}\:\mathrm{because}\:\mathrm{X}>\mathrm{0} \\ $$$$\mathrm{X}\:=\:\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:=\:\left(\frac{\mathrm{5}}{\mathrm{7}}\right)^{{x}} \\ $$$${x}\mathrm{ln}\frac{\mathrm{5}}{\mathrm{7}}\:=\:\mathrm{ln}\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right) \\ $$$${x}\:=\:\frac{\mathrm{ln}\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{ln}\frac{\mathrm{5}}{\mathrm{7}}} \\ $$

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