(3^2x)-(3^x) - 6 = 0?

3^(2x) - 3^x - 6 = 0
(3^x)^2 - 3^x - 6 = 0
Let y = 3^x
y^2 -y - 6 = 0
(y - 3)(y + 2) = 0
y = 3, -2
3^x = 3^1 => x = 1
3^x = -2 has no real solution.

Let y = 3^x

y ² - y - 6 = 0

( y - 3 ) ( y + 2 ) = 0

y = 3 ( accepting + ve value for y )

3^x = 3

x = 1

3^(2x) - 3^x - 6 = 0
3^x(3^x) - 3^x - 6 = 0

Let 3^x = y.

3^x(3^x) - 3^x - 6 = 0
y^2 - y - 6 = 0
y^2 + 2y - 3y - 6 = 0
(y^2 + 2y) - (3y + 6) = 0
y(y + 2) - 3(y + 2) = 0
(y + 2)(y - 3) = 0

y + 2 = 0
y = -2

y - 3 = 0
y = 3

Substitute 3^x for y.
3^x = -2
(extraneous solution since 3^x is not equal to a negative number)

3^x = 3
3^x = 3^1
x = 1

∴ x = 1

2xlog3-xlog3=6
xlog3=6
x=6/log3

Hi,
because 9-3-6=0
therefore 3^(2x)=9
3^2=9
x=1