anonymous
  • anonymous
Solve the equations using properties of logs: a.) 3log[11](x) = 2log[11]8 b.) 2log[5](x) = 3log[5]9
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
\[3\log_{11}(x)=2\log_{11}8 \]
anonymous
  • anonymous
From properties of logs\[3\log_{11}x=\log_{11}x^3\]and\[2\log_{11}8=\log_{11}8^2=\log_{11}64\]so the first equation boils down to\[\log_{11}x^3=\log_{11}64 \rightarrow x^3=64\]so that\[x=4\]
anonymous
  • anonymous
\[B.) 2\log_{5}(x)=3\log_{5}9 \]

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anonymous
  • anonymous
It's something similar for the second one too.
anonymous
  • anonymous
is B.) up there too?
anonymous
  • anonymous
?
anonymous
  • anonymous
there were two equations i posted, A and B
anonymous
  • anonymous
\[x^2=9^3=729 \rightarrow x=\pm 27\]
anonymous
  • anonymous
Yeah, I know, but the procedure is the same. The bases are different but it's irrelevant. It's just something to throw you off. As long as the bases are the same in your equation, you can remove the logs in the end using the definition of the log.
anonymous
  • anonymous
So the answer for B.) is + or - 27?
anonymous
  • anonymous
yes
anonymous
  • anonymous
where did the 729 come from?
anonymous
  • anonymous
9^3?
anonymous
  • anonymous
729 is 9^3

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