Question and Answers Forum

All Questions      Topic List

Logarithms Questions

Previous in All Question      Next in All Question      

Previous in Logarithms      Next in Logarithms      

Question Number 96636 by O Predador last updated on 03/Jun/20

Commented by hknkrc46 last updated on 03/Jun/20

5^x −(9,8)^x =7^x  ⇒ 5^x −(((49)/5))^x =7^x   ⇒ 5^x −(7^(2x) /5^x )=7^x  ⇒ 5^(2x) −7^(2x) =5^x ∙7^x   ⇒ ((5^(2x) −7^(2x) )/(5^x ∙7^x ))=1 ⇒ (5^x /7^x )−(7^x /5^x )=1  (5^x /7^x )=((5/7))^x =t ⇒ t−(1/t)=1 ⇒ t^2 −t−1=0  ⇒t=((−(−1)∓(√((−1)^2 −4∙1∙(−1))))/(2∙1))  ⇒((5/7))^x =((1∓(√5))/2) ⇒0 < ((5/7))^x = ((1+(√5))/2)  ⇒ln ((5/7))^x = ln ((1+(√5))/2)  ⇒ xln ((5/7))= ln ((1+(√5))/2) ⇒ x=((ln ((1+(√5))/2))/(ln (5/7)))

$$\mathrm{5}^{{x}} −\left(\mathrm{9},\mathrm{8}\right)^{{x}} =\mathrm{7}^{{x}} \:\Rightarrow\:\mathrm{5}^{{x}} −\left(\frac{\mathrm{49}}{\mathrm{5}}\right)^{{x}} =\mathrm{7}^{{x}} \\ $$$$\Rightarrow\:\mathrm{5}^{{x}} −\frac{\mathrm{7}^{\mathrm{2}{x}} }{\mathrm{5}^{{x}} }=\mathrm{7}^{{x}} \:\Rightarrow\:\mathrm{5}^{\mathrm{2}{x}} −\mathrm{7}^{\mathrm{2}{x}} =\mathrm{5}^{{x}} \centerdot\mathrm{7}^{{x}} \\ $$$$\Rightarrow\:\frac{\mathrm{5}^{\mathrm{2}{x}} −\mathrm{7}^{\mathrm{2}{x}} }{\mathrm{5}^{{x}} \centerdot\mathrm{7}^{{x}} }=\mathrm{1}\:\Rightarrow\:\frac{\mathrm{5}^{{x}} }{\mathrm{7}^{{x}} }−\frac{\mathrm{7}^{{x}} }{\mathrm{5}^{{x}} }=\mathrm{1} \\ $$$$\frac{\mathrm{5}^{{x}} }{\mathrm{7}^{{x}} }=\left(\frac{\mathrm{5}}{\mathrm{7}}\right)^{{x}} ={t}\:\Rightarrow\:{t}−\frac{\mathrm{1}}{{t}}=\mathrm{1}\:\Rightarrow\:{t}^{\mathrm{2}} −{t}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{t}=\frac{−\left(−\mathrm{1}\right)\mp\sqrt{\left(−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}\centerdot\mathrm{1}\centerdot\left(−\mathrm{1}\right)}}{\mathrm{2}\centerdot\mathrm{1}} \\ $$$$\Rightarrow\left(\frac{\mathrm{5}}{\mathrm{7}}\right)^{{x}} =\frac{\mathrm{1}\mp\sqrt{\mathrm{5}}}{\mathrm{2}}\:\Rightarrow\mathrm{0}\:<\:\left(\frac{\mathrm{5}}{\mathrm{7}}\right)^{{x}} =\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{ln}\:\left(\frac{\mathrm{5}}{\mathrm{7}}\right)^{{x}} =\:\mathrm{ln}\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\Rightarrow\:{x}\mathrm{ln}\:\left(\frac{\mathrm{5}}{\mathrm{7}}\right)=\:\mathrm{ln}\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\Rightarrow\:{x}=\frac{\mathrm{ln}\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}}{\mathrm{ln}\:\frac{\mathrm{5}}{\mathrm{7}}} \\ $$

Answered by mr W last updated on 03/Jun/20

((1/(1.4)))^x −(1.4)^x =1  (1.4)^(2x) +(1.4)^x −1=0  1.4^x =(((√5)−1)/2)  ⇒x=((ln ((√5)−1)−ln 2)/(ln 7−ln 5))≈−1.4302

$$\left(\frac{\mathrm{1}}{\mathrm{1}.\mathrm{4}}\right)^{{x}} −\left(\mathrm{1}.\mathrm{4}\right)^{{x}} =\mathrm{1} \\ $$$$\left(\mathrm{1}.\mathrm{4}\right)^{\mathrm{2}{x}} +\left(\mathrm{1}.\mathrm{4}\right)^{{x}} −\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{1}.\mathrm{4}^{{x}} =\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{ln}\:\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)−\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{7}−\mathrm{ln}\:\mathrm{5}}\approx−\mathrm{1}.\mathrm{4302} \\ $$

Commented by O Predador last updated on 03/Jun/20

 or  log_(7/5) (((−1+(√5))/2))  ?

$$\:\boldsymbol{\mathrm{or}}\:\:\boldsymbol{\mathrm{log}}_{\frac{\mathrm{7}}{\mathrm{5}}} \left(\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\:\:? \\ $$

Commented by mr W last updated on 03/Jun/20

yes

$${yes} \\ $$

Commented by O Predador last updated on 03/Jun/20

 Thank you!

$$\:\boldsymbol{\mathrm{Thank}}\:\boldsymbol{\mathrm{you}}! \\ $$

Answered by 1549442205 last updated on 03/Jun/20

Equation is equivalent to  5^x −(((49)/5))^x =7^x ⇔5^(2x) −7^(2x) =7^x ×5^x (1)  ⇔((5/7))^x −((7/5))^x =1  et ((7/5))^x =y(y>0)  ⇒(1/y)−y=1⇔y^2 +y−1=0  ⇒y=((−1+(√5))/2)⇒((7/5))^t =((−1+(√5))/2)  ⇒x=((ln(((−1+(√5))/2)))/(ln(7/5)))≈−1.43016

$$\mathrm{Equation}\:\mathrm{is}\:\mathrm{equivalent}\:\mathrm{to} \\ $$$$\mathrm{5}^{\mathrm{x}} −\left(\frac{\mathrm{49}}{\mathrm{5}}\right)^{\mathrm{x}} =\mathrm{7}^{\mathrm{x}} \Leftrightarrow\mathrm{5}^{\mathrm{2x}} −\mathrm{7}^{\mathrm{2x}} =\mathrm{7}^{\mathrm{x}} ×\mathrm{5}^{\mathrm{x}} \left(\mathrm{1}\right) \\ $$$$\Leftrightarrow\left(\frac{\mathrm{5}}{\mathrm{7}}\right)^{\mathrm{x}} −\left(\frac{\mathrm{7}}{\mathrm{5}}\right)^{\mathrm{x}} =\mathrm{1} \\ $$$$\mathrm{et}\:\left(\frac{\mathrm{7}}{\mathrm{5}}\right)^{\mathrm{x}} =\mathrm{y}\left(\mathrm{y}>\mathrm{0}\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{y}}−\mathrm{y}=\mathrm{1}\Leftrightarrow\mathrm{y}^{\mathrm{2}} +\mathrm{y}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{y}=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\Rightarrow\left(\frac{\mathrm{7}}{\mathrm{5}}\right)^{\mathrm{t}} =\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{x}=\frac{\mathrm{ln}\left(\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)}{\mathrm{ln}\frac{\mathrm{7}}{\mathrm{5}}}\approx−\mathrm{1}.\mathrm{43016} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com