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Question Number 167231 by mathlove last updated on 10/Mar/22

Commented by cortano1 last updated on 10/Mar/22

2^x .5^(1/x) = 10 log _(10) (2^x .5^(1/x) )=1 log _(10) (2^x )+log _(10) (5^(1/x) )=1 x.log _(10) (2)+(1/x)log _(10) (5)=1 (x−1)log _(10) (2)=(1−(1/x))log _(10) (5) ((x−1)/((((x−1)/x)))) = ((log _(10) (5))/(log _(10) (2))) x = log _2 (5)

$$\:\:\mathrm{2}^{\mathrm{x}} .\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{x}}} =\:\mathrm{10} \\ $$$$\:\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{2}^{\mathrm{x}} .\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{x}}} \right)=\mathrm{1} \\ $$$$\:\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{2}^{\mathrm{x}} \right)+\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{x}}} \right)=\mathrm{1} \\ $$$$\:\mathrm{x}.\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{x}}\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{5}\right)=\mathrm{1} \\ $$$$\:\left(\mathrm{x}−\mathrm{1}\right)\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{2}\right)=\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}}\right)\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{5}\right) \\ $$$$\:\frac{\mathrm{x}−\mathrm{1}}{\left(\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}}\right)}\:=\:\frac{\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{5}\right)}{\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{2}\right)} \\ $$$$\:\mathrm{x}\:=\:\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{5}\right) \\ $$

Answered by Rasheed.Sindhi last updated on 10/Mar/22

2^x ∙5^((1+x)/x) =50 Assuming x a whole number: 2^x ∙5^((1+x)/x) =2^1 ∙5^2 x=1∧ ((1+x)/x)=2⇒x=1

$$\mathrm{2}^{{x}} \centerdot\mathrm{5}^{\frac{\mathrm{1}+{x}}{{x}}} =\mathrm{50} \\ $$$${Assuming}\:{x}\:{a}\:{whole}\:{number}: \\ $$$$\mathrm{2}^{{x}} \centerdot\mathrm{5}^{\frac{\mathrm{1}+{x}}{{x}}} =\mathrm{2}^{\mathrm{1}} \centerdot\mathrm{5}^{\mathrm{2}} \\ $$$${x}=\mathrm{1}\wedge\:\frac{\mathrm{1}+{x}}{{x}}=\mathrm{2}\Rightarrow{x}=\mathrm{1} \\ $$

Answered by mr W last updated on 10/Mar/22

2^x ×5×5^(1/x) =50 2^x ×5^(1/x) =10 2^(x−1) =5^(1−(1/x)) (x−1)ln 2=(((x−1)/x))ln 5 (x−1)(ln 2−((ln 5)/x))=0 ⇒x−1=0 ⇒x=1 ⇒ln 2−((ln 5)/x)=0 ⇒x=((ln 5)/(ln 2))

$$\mathrm{2}^{{x}} ×\mathrm{5}×\mathrm{5}^{\frac{\mathrm{1}}{{x}}} =\mathrm{50} \\ $$$$\mathrm{2}^{{x}} ×\mathrm{5}^{\frac{\mathrm{1}}{{x}}} =\mathrm{10} \\ $$$$\mathrm{2}^{{x}−\mathrm{1}} =\mathrm{5}^{\mathrm{1}−\frac{\mathrm{1}}{{x}}} \\ $$$$\left({x}−\mathrm{1}\right)\mathrm{ln}\:\mathrm{2}=\left(\frac{{x}−\mathrm{1}}{{x}}\right)\mathrm{ln}\:\mathrm{5} \\ $$$$\left({x}−\mathrm{1}\right)\left(\mathrm{ln}\:\mathrm{2}−\frac{\mathrm{ln}\:\mathrm{5}}{{x}}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}−\mathrm{1}=\mathrm{0}\:\Rightarrow{x}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{ln}\:\mathrm{2}−\frac{\mathrm{ln}\:\mathrm{5}}{{x}}=\mathrm{0}\:\Rightarrow{x}=\frac{\mathrm{ln}\:\mathrm{5}}{\mathrm{ln}\:\mathrm{2}} \\ $$

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