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 $\displaystyle 3^{x + 2} = 3^x\cdot 3^2 = 9\left(3^x\right)$.

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

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

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.

$\displaystyle 3^x = 8$

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

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

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

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

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