3^(2x) - 3^(x+2) +8=0
i tried 3^(x)=y but what about the 3^(x+2) ?
Notice that $\displaystyle 3^{x + 2} = 3^x\cdot 3^2 = 9\left(3^x\right)$.
So the equation becomes
$\displaystyle 3^{2x} - 9\left(3^x\right) + 8 = 0$.
Also notice that $\displaystyle 3^{2x} = \left(3^x\right)^2$.
So the equation becomes
$\displaystyle \left(3^x\right)^2 - 9\left(3^x\right) + 8 = 0$.
This is a Quadratic Equation.
Bring in a dummy variable $\displaystyle X = 3^x$.
So the equation becomes
$\displaystyle X^2 - 9X + 8 = 0$
$\displaystyle X^2 - 8X - X + 8 = 0$
$\displaystyle X(X - 8) - 1(X - 8) = 0$
$\displaystyle (X - 1)(X - 8) = 0$.
Therefore $\displaystyle X - 1 = 0$ or $\displaystyle X - 8 = 0$
$\displaystyle X = 1$ or $\displaystyle X = 8$.
So $\displaystyle 3^x = 1$ or $\displaystyle 3^x = 8$
$\displaystyle x = 0$ or $\displaystyle x = \log_3{8} = \log_3{2^3} = 3\log_3{2}$.
You could also work out $\displaystyle x$ using natural logarithms.
$\displaystyle 3^x = 8$
$\displaystyle \ln{3^x} = \ln{8}$
$\displaystyle x\ln{3} = \ln{2^3}$
$\displaystyle x\ln{3} = 3\ln{2}$
$\displaystyle x = \frac{3\ln{2}}{\ln{3}}$.