anonymous
  • anonymous
Algebra 2 help would be greatly appreciated!
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
\[3^{2x}-15(3^{x})+56=0\]
Astrophysics
  • Astrophysics
Hey so what's it asking
Astrophysics
  • Astrophysics
Solve for x? :O

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anonymous
  • anonymous
Yeah solve for x
madhu.mukherjee.946
  • madhu.mukherjee.946
substitute 3^x as a and solve the equation as a quadratic equation
madhu.mukherjee.946
  • madhu.mukherjee.946
do you need more explanation ?i mean can do it on your own
madhu.mukherjee.946
  • madhu.mukherjee.946
anonymous
  • anonymous
can i multiply the -15 and 3^x ???
Astrophysics
  • Astrophysics
No but you can simplify it -5(3^x+1)
Astrophysics
  • Astrophysics
\[-5(3^{x+1})\]
anonymous
  • anonymous
don't seem to be quite understanding this :/
Astrophysics
  • Astrophysics
You have to know your exponent rules, 5 x 3 = 15, so we have 5(3)(3^x) you can only do this with like bases so we have \[3^{x+1}\]
Astrophysics
  • Astrophysics
So here is the general rule \[x^mx^n = x^{m+n}\]
Astrophysics
  • Astrophysics
You can factor this I suppose
Astrophysics
  • Astrophysics
Yes, factoring should work
Astrophysics
  • Astrophysics
factoring you should end up with \[(3^x-8)(3^x-7)=0\] as we are treating 3^x as a, also suggested above
anonymous
  • anonymous
ohh this makes a little more sense now!
Astrophysics
  • Astrophysics
-7 x - 8 = 56 -7+-8=15
ganeshie8
  • ganeshie8
\[\large 3^{2x}-15(3^{x})+56=0\] \[\large (\color{red}{3^{x}})^2-15(\color{red}{3^{x}})+56=0\] let \(t=\color{red}{3^{x}}\) \[\large (\color{red}t{})^2-15(\color{red}{t})+56=0\] you knw how to handle a quadratic
Astrophysics
  • Astrophysics
-15*
anonymous
  • anonymous
Yes!! Thanks everyone for helping! Really appreciate it.
Astrophysics
  • Astrophysics
You can solve for x now, it involves logarithms, you know how to do that right? \[3^x - 7 = 0\] set it to 0 and solve for x for both
anonymous
  • anonymous
Yeah of course. Thanks again everyone!
Astrophysics
  • Astrophysics
Ok, yw :)

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