anonymous
  • anonymous
Solve for x (IB HL Math question)
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
\[\Large \ln(x)+\log_{x}\sqrt{e}=\frac{1}{2}\]
anonymous
  • anonymous

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dan815
  • dan815
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dan815
  • dan815
|dw:1438304890557:dw|
anonymous
  • anonymous
Oh I figured it out. Change of base formula
anonymous
  • anonymous
I aso wrote the wrong problem. This one gives imag solutions \[\Large \ln(x)+\log_{x}\sqrt{e}=\frac{3}{2}\]
dan815
  • dan815
okay i see
dan815
  • dan815
ya apply change of base and u can play around with it
anonymous
  • anonymous
So \[\Large 2\ln(x)+\frac{\ln(e)}{\ln(x)}=3\]
dan815
  • dan815
yeah
dan815
  • dan815
multiply everything by ln(x) and set y = ln(x) solve for y for the quadratic u get then solve for x
anonymous
  • anonymous
and then \[\Large 2(lnx)^{2}+1=3\ln(x)\] If we set (lnx)=x \[\Large 2(x)^2-3(x)+1=0\]
anonymous
  • anonymous
Quad formula gives us x=1 and x=1/2
anonymous
  • anonymous
So pluggin x back in gives us lnx=1 and lnx=1/2 x= e and x= sqrt(e)
dan815
  • dan815
yes
dan815
  • dan815
plug in your answers to check
anonymous
  • anonymous
ln(e)+log_e{sqrt{e}}= 1 + 1/2 = 3/2
anonymous
  • anonymous
ln(sqrt(e)}+log_sqrt(e)(sqrt(e)}= 1/2 +1 = 3/2

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