Solving an exponential equation of quadratic form

**smmmc** · January 8th 2010, 08:25 PM

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

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

**Prove It** · January 8th 2010, 09:04 PM

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

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

This is a Quadratic Equation.

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.

$3^x = 8$

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

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

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

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

Thread: Solving an exponential equation of quadratic form

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