anonymous
  • anonymous
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Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
ok you are thinking that \[3^{2x}=(3^x)^2\] so if you put \[z=3^x\] you have a quadratic equation \[z^2=5z+36\] \[z^2-5z-36=0\] \[(z-9)(z+3)=0\] \[z=9,z=-3\]
anonymous
  • anonymous
now replace z by 3^x get \[3^x=9,3^x=-3\]
anonymous
  • anonymous
you can forget about \[3^x=-3\] because no matter what x is, \[3^x>0\] so it cannot be -3

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anonymous
  • anonymous
as for \[3^x=9\] it is clear that \[3^2=9\] meaning \[x=2\]
anonymous
  • anonymous
forget the damned logs, they are not important, and i see i made a small mistake, it should be \[(z-9)(z+4)=0\] but the result is the same
anonymous
  • anonymous
Wow that actually makes so much more sense! Thanks @satellite73 ! The logs were confusing me
anonymous
  • anonymous
you do not need logs to solve \[3^x=9\] or \[3^x=27\] you might need them later for more difficult problems
anonymous
  • anonymous
Alright. thanks again @satellite73, you explanation makes so much more sense
anonymous
  • anonymous
@satellite73, how does it go from z2āˆ’5zāˆ’36=0 to (zāˆ’9)(z+3)=0? what happens here?
anonymous
  • anonymous
that was a typo i corrected it should be \[z^2-5z-36=(z-9)(z+4)\]
anonymous
  • anonymous
Oh okok. Thank you!

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