How do you solve $3^(2x+1) - 26(3^x) - 9 = 0$ ?

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Answer 1 · 2016-04-07T10:42:52

$x=2$

Replace $3^x$ by $y$ and you will obtain:

$3y^2-26y-9$ which can be solved using the quadratic formula:

$y=(26+-sqrt(26^2-4*3*(-9)))/(2*3)$

$y=(26+-sqrt(676+108))/(6)$

$y=(26+-sqrt(784))/(6)$

$y=(26+-28)/(6)$

$y=54/6=9 or y=-2/6=-1/3$

know we solve:

$3^x=9$ or $3^x=-1/3$

$3^x=3^2$ or $no solution$

$x=2$

How do you solve 3^(2x+1) - 26(3^x) - 9 = 03^(2x+1) - 26(3^x) - 9 = 0?