anonymous
  • anonymous
What is the value of x in the equation.... (logarithms), equation will be posted in comments
Algebra
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\log_{8}x + \log_{4}x = 10/3 \]
kc_kennylau
  • kc_kennylau
Firstly, the law: \[\Large\log_ab=\dfrac{\log_nb}{\log_na}\] Substitute \(\Large a=8\), \(\Large b=x\) to get: \[\Large\log_8x=\dfrac{\log_nx}{\log_n8}\]
kc_kennylau
  • kc_kennylau
Substitute \(\Large a = 4\), \(\Large b = x\) to get: \[\Large\log_4x=\dfrac{\log_nx}{\log_n4}\]

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kc_kennylau
  • kc_kennylau
What would be a reasonable \(\Large n\) to fit in?
anonymous
  • anonymous
10?
kc_kennylau
  • kc_kennylau
\[\Large \log_8x+\log_4x=\dfrac{10}3\] \[\Large \dfrac{\log_nx}{\log_n8}+\dfrac{\log_nx}{\log_n4}=\dfrac{10}3\] I would actually choose \(\Large n = 2\) because \(\Large 4=2^2\) and \(\Large 8=2^3\)
kc_kennylau
  • kc_kennylau
Let's use \(\Large n = 10\) anyway.
kc_kennylau
  • kc_kennylau
\[\Large \dfrac{\log_{10}x}{\log_{10}2^3}+\dfrac{\log_{10}x}{\log_{10}2^2}=\dfrac{10}3\]
anonymous
  • anonymous
I was debating whether to use 2, but I didn't know how to use the 2 to the power of 2 and 3 in the equation.
kc_kennylau
  • kc_kennylau
okay, let's use \(\Large n = 2\) then.
kc_kennylau
  • kc_kennylau
\[\Large \dfrac{\log_2x}{\log_22^3}+\dfrac{\log_2x}{\log_22^2}=\dfrac{10}3\]
kc_kennylau
  • kc_kennylau
\[\Large \dfrac{\log_2x}3+\dfrac{\log_2x}2=\dfrac{10}3\]
kc_kennylau
  • kc_kennylau
Are you able to find \(\Large \log_2x\) now?
anonymous
  • anonymous
Sorry, I'm not completely sure how to finish it.
anonymous
  • anonymous
Would we simplify them into one fraction?
kc_kennylau
  • kc_kennylau
Let's substitute \(\Large a \) as \(\Large \log_2x\).
kc_kennylau
  • kc_kennylau
The equation then becomes: \[\Large\frac a3+\frac a2=\frac{10}3\]
anonymous
  • anonymous
a=4?
kc_kennylau
  • kc_kennylau
Yes :)
anonymous
  • anonymous
x=16!
kc_kennylau
  • kc_kennylau
So \(\Large \log_2x=4\)
kc_kennylau
  • kc_kennylau
Yes :D
kc_kennylau
  • kc_kennylau
Smart boy/girl
anonymous
  • anonymous
Haha, girl - Thank you so much!
kc_kennylau
  • kc_kennylau
no problem :pp

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