Solve 3^(2x+1) - 28 (3^x-1) +1=0?

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Answer 1 · 2018-04-04T10:06:34

#x=1 or -2#

Split the addition and subtraction of exponents into multiplication and division.

#3^2x x 3^1 -28(3^x /3^1) + 1=0#

Let

#u=3^x#

Now

#3u^2 -28(u/3) +1=0#

#3u^2 - 28u/3 +1 =0#

#9u^2 -28u +3=0#

Using the quadratic equation

#u_(1,2) = (-b +- sqrt(b^2 – 4ac))/(2a)#

you get

#u = (28 +- sqrt(28^2 -4xx9xx3))/(2xx9)#

#u = (28 +- sqrt(784-108))/18#

#u = (28 +- sqrt(676))/18#

#u = (28 +- 26)/18#

#u= 54/18 or u = 2/18#

#u=3 or u = 1/9#

Put #3^x# back in for #u# to get

#3^x=3 or 3^x=1/9#

#3^x=3^1 or 3^x=3^-2#

Therefore,

#x=1 or x = -2#