villy9mili7acieniqu
villy9mili7acieniqu villy9mili7acieniqu
  • 12-07-2016
  • Mathematics
contestada

what is the true solution to the equation below?
2 ln e^(ln 2x)-ln e^(ln 10x)=ln 30
x=30
x=75
x150
x=300

Respuesta :

dalendrk
dalendrk dalendrk
  • 13-07-2016
[tex]D:x\in\mathbb{R^+}\\\\2\ln e^{\ln2x}-\ln e^{\ln10x}=\ln30\\2\ln2x\cdot\ln e-\ln10x\cdot\ln e=\ln30\\2\ln2x-\ln10x=\ln30\\\ln(2x)^2-\ln10x=\ln30\\\ln\dfrac{4x^2}{10x}=\ln30\\\ln\dfrac{2x}{5}=\ln30\iff\dfrac{2}{5}x=30\ \ \ \multiply\ both\ sides\ by\ \dfrac{5}{3}\\\boxed{x=75}\in D[/tex]
Answer Link
shermanmakayla1677
shermanmakayla1677 shermanmakayla1677
  • 12-12-2018

Answer:

x=75

Step-by-step explanation:


Answer Link

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