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So I have been dealing with an extremely difficult equation: $x^{2a} = \frac{x}{a}$ and am confused on how to solve it. I am wondering how to solve for $x $ in terms of $ a$ in the case of this question. If possible, give an explanation of the answer.

Martund
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James
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3 Answers3

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I am assuming you want to solve: $$x^{2a} = \frac{x}{a}\implies 2a\ln(x)=\ln(x)-\ln(a)\implies \ln(x)=\frac{\ln(a)}{1-2a}$$ $$\implies x=e^{\frac{\ln(a)}{1-2a}} = \left(e^{\ln a}\right)^{{1/(1-2a)}}=a^{\frac{1}{1-2a}}.$$

Another way would be to do:

$$x^{2a} = \frac{x}{a}\implies x^{2a-1}=a^{-1}\implies x=a^{\frac{1}{1-2a}}.$$

Student
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$$x^{2a}=x/a$$You can divide both sides by $x$ to get $$\frac{x^{2a}}{x}=\frac1a$$Use index rules to get $$x^{2a-1}=\frac1a$$Then raise both sides to the power $\frac1{2a-1}$ to get $$x=\left(\frac{1}a\right)^{\frac1{2a-1}}=a^{\frac1{1-2a}}$$

John Doe
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  • How can Hello World's answer be simplified to your answer? – James Jan 06 '19 at 05:03
  • $$e^{\frac{\ln a}{1-2a}}=(e^{\ln a})^{\frac1{1-2a}}=a^{\frac1{1-2a}}$$Since $e^{\ln a}=a$ by the fact that these are inverse functions of each other. – John Doe Jan 06 '19 at 05:04
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This may be the way to proceed: Hope this helps. .