# How do you solve 3^(2x) – 2*3^(x+5) + 3^10 = 0?

May 18, 2016

$x = 5$

#### Explanation:

Remember the polynomial identity
${\left(a + b\right)}^{2} = {a}^{2} + 2 a x + {b}^{2}$. Choosing $a = {3}^{x}$ and $b = - {3}^{5}$ we have
${\left({3}^{x} - {3}^{5}\right)}^{2} = {3}^{2 x} - 2 \cdot {3}^{x} \cdot {3}^{5} + {3}^{10}$. So, our problem is reduced to: Solve for $x$ the condition ${\left({3}^{x} - {3}^{5}\right)}^{2} = 0 \to {3}^{x} - {3}^{5} = 0 \to {3}^{x} = {3}^{5} \to x = 5$

May 18, 2016

x=5

#### Explanation:

At first sight this seems a quadratic equation of the form:
${y}^{2} + b y + c$ where $y = {3}^{x}$. So let's transform the equation into the quadratic form:

${3}^{2 x} - 2 \times {3}^{5} \times {3}^{x} + {3}^{10}$

We can see that the equation can also be reduced to

${\left({3}^{x}\right)}^{2} - 2 \times {3}^{5} \times {3}^{x} + {\left({3}^{5}\right)}^{2}$

Remember that :

$\textcolor{b l u e}{{a}^{2} - 2 a b + {b}^{2} = {\left(a - b\right)}^{2}}$

The equation can be arranged into:

${\left({3}^{x} - {3}^{5}\right)}^{2} = 0$

So:

${3}^{x} - {3}^{5} = 0$

${3}^{x} = {3}^{5}$

$x = 5$