anonymous
  • anonymous
30=50(1-0.9^x)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
are u solving for x?
anonymous
  • anonymous
yes
anonymous
  • anonymous
30= (50 X 1 X -0.9^x)

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More answers

anonymous
  • anonymous
i tried to solve it but you can't take the log of a negative to get x
anonymous
  • anonymous
30= 50 - 45^x
anonymous
  • anonymous
why would you distribute the 50
DebbieG
  • DebbieG
Hmmm.... 30=50(1-0.9^x) 3/5=1-0.9^x 0.9^x=2/5 Then switch to log form: \[\Large \log_{0.9} \frac{ 2 }{ 5 }=x\] You'll need the change of base formula to get a numerical result.
anonymous
  • anonymous
okay. that works. Makes sense! thanks!
DebbieG
  • DebbieG
You can't distribute the 50 like that. PEMDAS... exponents before multiplication. At best, you can: 30=50(1-0.9^x) 30=50-50(0.9^x) But that doesn't really help matters. :)
DebbieG
  • DebbieG
You're welcome. :)
campbell_st
  • campbell_st
you don't need change of base... apply the log law for powers xln(0.9) = ln(2/5) solve for x
anonymous
  • anonymous
or you could use a calculator and do log(2/5)/log(0.9)
DebbieG
  • DebbieG
Yup, that works too.... just depends on what methods you've learned and prefer. It's really the same thing. :)
DebbieG
  • DebbieG
@fullib , that is the change of base formula. :) You can either use Log or Ln.
DebbieG
  • DebbieG
Notice that from: xln(0.9) = ln(2/5) You get: x=ln(2/5) / ln(0.9) Change of base says that; \[\Large \log_{0.9} \frac{ 2 }{ 5 }=\frac{ \ln(2/5) }{ \ln0.9 }\] So they are really just two slightly different routes to the exact same result. :)
campbell_st
  • campbell_st
so make the solution easier... and minimising mistake by eliminating unnecessary steps

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