3^(2x)-5(3^x)+4=0 , Find X?

3^(2x) = (3^x)²

Let u = 3^x

u²-5u+4 = 0
(u-4)(u-1) = 0

u = {1,4}
3^x = {1, 4}

x = {0, ln(4)/ln(3)}

3^(2x) - 5(3^x) + 4 = 0
(3^x)^2 - 5(3^x) + 4 = 0

Change 3^x into y:
(3^x)^2 - 5(3^x) + 4 = 0
y^2 - 5y + 4 = 0
y^2 - y - 4y + 4 = 0
(y^2 - y) - (4y - 4) = 0
y(y - 1) - 4(y - 1) = 0
(y - 1)(y - 4) = 0

y - 1 = 0
y = 1

y - 4 = 0
y = 4

Change y into 3^x:
y = 1
3^x = 1
3^x = 3^0
x = 0

y = 4
3^x = 4
x = log_3(4)

∴ x = 0 , log_3(4)

Let y = 3^x

Then
3^(2x) - 5(3^x) + 4 = y^2 - 5y + 4 = 0

(y - 4)(y - 1) = 0

So y = 4 and y = 1

substitute y = 3^x back in to get

3^x = 4 and 3^x = 1

now take logs

3^x = 4
log(3^x) = log(4)
xlog(3) = log(4)
x = log(4)/log(3) = 2[log(2)/log(3)]

3^x = 1
log(3^x) = log(1) = 0
xlog(3) = 0
x = 0

Hope that helps :)