# Math Help - Solving an exponential equation of quadratic form

1. 3^(2x) - 3^(x+2) +8=0

i tried 3^(x)=y but what about the 3^(x+2) ?

2. Originally Posted by smmmc
3^(2x) - 3^(x+2) +8=0

i tried 3^(x)=y but what about the 3^(x+2) ?
Notice that $3^{x + 2} = 3^x\cdot 3^2 = 9\left(3^x\right)$.

So the equation becomes

$3^{2x} - 9\left(3^x\right) + 8 = 0$.

Also notice that $3^{2x} = \left(3^x\right)^2$.

So the equation becomes

$\left(3^x\right)^2 - 9\left(3^x\right) + 8 = 0$.

Bring in a dummy variable $X = 3^x$.

So the equation becomes

$X^2 - 9X + 8 = 0$

$X^2 - 8X - X + 8 = 0$

$X(X - 8) - 1(X - 8) = 0$

$(X - 1)(X - 8) = 0$.

Therefore $X - 1 = 0$ or $X - 8 = 0$

$X = 1$ or $X = 8$.

So $3^x = 1$ or $3^x = 8$

$x = 0$ or $x = \log_3{8} = \log_3{2^3} = 3\log_3{2}$.

You could also work out $x$ using natural logarithms.

$3^x = 8$

$\ln{3^x} = \ln{8}$

$x\ln{3} = \ln{2^3}$

$x\ln{3} = 3\ln{2}$

$x = \frac{3\ln{2}}{\ln{3}}$.