chika_tika3007
2^2x 6.2^x +8 =0



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ganeshie8
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1
say k = \(\large 2^x\)

ganeshie8
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\(\large 2^{2x} 6.2^x +8 =0\)
\(\large (2^{x})^2 6.2^x +8 =0\)

ganeshie8
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plugin k and you will see a quadratic in k
solve k

chika_tika3007
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ok but how abaout the number 8???

ganeshie8
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wat do u get after putting \(2^x=k\)
?

ganeshie8
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we get this :
\(\large k^26k+8=0\)

ganeshie8
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this is just a quadratic equation
you knw how to solve a quadratic equaiton ha ?

chika_tika3007
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yes

ganeshie8
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good, factor it and solve k first

chika_tika3007
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humm ok ok i got it k thanks

ganeshie8
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sure ?

ganeshie8
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\(\large (k2)(k4) = 0\)
\(k = 2, k=4\)

chika_tika3007
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i get the answer are 2 and 4

chika_tika3007
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yes..

ganeshie8
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when k = 2,
\(2^x = k\)
\(2^x = 2^1\)
\(x = 1\)

ganeshie8
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when k = 4,
\(2^x = k\)
\(2^x = 4\)
\(2^x = 2^2\)
\(x = 2\)

ganeshie8
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so the solutions are : 1, 2

chika_tika3007
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yup yup

ganeshie8
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cool :)