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Linear Algebra
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2^2x -6.2^x +8 =0
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say k = \(\large 2^x\)
\(\large 2^{2x} -6.2^x +8 =0\) \(\large (2^{x})^2 -6.2^x +8 =0\)
plugin k and you will see a quadratic in k solve k
ok but how abaout the number 8???
wat do u get after putting \(2^x=k\) ?
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we get this :- \(\large k^2-6k+8=0\)
this is just a quadratic equation you knw how to solve a quadratic equaiton ha ?
yes
good, factor it and solve k first
humm ok ok i got it k thanks
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sure ?
\(\large (k-2)(k-4) = 0\) \(k = 2, k=4\)
i get the answer are -2 and -4
yes..
when k = 2, \(2^x = k\) \(2^x = 2^1\) \(x = 1\)
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when k = 4, \(2^x = k\) \(2^x = 4\) \(2^x = 2^2\) \(x = 2\)
so the solutions are : 1, 2
yup yup
cool :)
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