anonymous
  • anonymous
solve: log(underscore:x)9 = 1/2
Mathematics
schrodinger
  • schrodinger
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lgbasallote
  • lgbasallote
_ <--underscore (shift + minus sign)
lgbasallote
  • lgbasallote
anyway... \[\log_x 9 = \frac 12\]
lgbasallote
  • lgbasallote
right?

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anonymous
  • anonymous
yes igbasallote
anonymous
  • anonymous
that's the equation
lgbasallote
  • lgbasallote
\[\log_x 9 = \frac 12\] first step is to change it into exponential form \[x^{1/2} = 9\] do you agree with this form
anonymous
  • anonymous
yes
lgbasallote
  • lgbasallote
good. so we need x. you have x^1/2 what do you think you should do to make that x^1/2 into just x^1?
anonymous
  • anonymous
could you log both sides, and bring down the 1/2
lgbasallote
  • lgbasallote
hmm not exactly
lgbasallote
  • lgbasallote
here's a hint \[\LARGE (a^3)^{1/3} = a^{3/3} = a^1 = a\] \[\LARGE (a^2)^{1/2} = a^{2/2} = a^1 = a\] \[\LARGE (a^{1/3})^3 = a^{3/3} = a^1 = a\]
anonymous
  • anonymous
oh, do you square both sides?
lgbasallote
  • lgbasallote
yup
anonymous
  • anonymous
so x = 81?
lgbasallote
  • lgbasallote
yup
anonymous
  • anonymous
perfect thanks!

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