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

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Answer 1 · 2018-04-04T10:06:34

$x = 1 or - 2$ $x=1 or -2$

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

$3^{2} x x 3^{1} - 28 (\frac{3^{x}}{3^{1}}) + 1 = 0$ $3^2x x 3^1 -28(3^x /3^1) + 1=0$

Let

$u = 3^{x}$ $u=3^x$

Now

$3 u^{2} - 28 (\frac{u}{3}) + 1 = 0$ $3u^2 -28(u/3) +1=0$

$3 u^{2} - 28 \frac{u}{3} + 1 = 0$ $3u^2 - 28u/3 +1 =0$

$9 u^{2} - 28 u + 3 = 0$ $9u^2 -28u +3=0$

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,

$x=1 or x = -2$