# If 216 * 3^(2x)-73*3^x-3=0 and u=3^x, what is the value of u?

Oct 5, 2017

$u = 0.375$

#### Explanation:

If $\textcolor{b l u e}{u} = \textcolor{m a \ge n t a}{{3}^{x}}$
then
$\textcolor{w h i t e}{\text{xxx}} 216 \cdot \textcolor{m a \ge n t a}{{3}^{\textcolor{b l a c k}{2} x}} + \left(- 73\right) \cdot \textcolor{m a \ge n t a}{{3}^{x}} - 3 = 0$
becomes
$\textcolor{w h i t e}{\text{xxx}} 216 {\textcolor{b l u e}{u}}^{2} - 73 \textcolor{b l u e}{u} - 3 = 0$

Using the quadratic formula (and a calculator)
$\textcolor{w h i t e}{\text{xxx}} \textcolor{b l u e}{u} = \frac{73 \pm \sqrt{{\left(- 73\right)}^{2} - 4 \cdot 216 \cdot \left(- 3\right)}}{2 \cdot 216}$

$\textcolor{w h i t e}{\text{xxx}} = 0.375$
or
$\textcolor{w h i t e}{\text{xxx}} = - 0.037 \overline{037}$

...but if $u = {3}^{x}$ then $u > 0 , \forall x \in \mathbb{R}$
and therefore the negative version is extraneous,
leaving only
$\textcolor{w h i t e}{\text{xxx}} \textcolor{b l u e}{u} = 0.375$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

As a bonus we could find the value of $x$ (approximately)

If ${3}^{x} = u$
then
$\textcolor{w h i t e}{\text{xxx}} {\log}_{3} \left({3}^{x}\right) = {\log}_{3} \left(u\right)$

$\textcolor{w h i t e}{\text{xxx}} x = {\log}_{3} \left(u\right)$

$\textcolor{w h i t e}{\text{xxx}} x = {\log}_{3} \left(0.375\right)$

then using a calculator
$\textcolor{w h i t e}{\text{xxx}} x \approx - 0.892789261$

Oct 5, 2017

$x = - 0.89$

#### Explanation:

The equation is in the form, $a {u}^{2} - b u - c$ where $u = {3}^{x}$.
Since it is difficult to use $x$ straightaway to solve the equation, we have replaced it with ‘u’.
Once we solve for u, we will then replace u with 3^x and find the value of x.

$216 {u}^{2} - 73 u - 3 = 0$
$216 \cdot 3 = 648$ can be factorised as $81 \cdot 8$
$216 {u}^{2} + 8 u - 81 u - 3 = 0$
$8 u \cdot \left(27 u + 1\right) - 3 \cdot \left(27 u + 1\right) = 0$
$\left(27 u + 1\right) \left(8 u - 3\right) = 0$
u=-(1/27) & (3/8)

${3}^{x} = - \left(\frac{1}{27}\right) = - {3}^{-} 3$ This solution is imaginary.

${3}^{x} = \frac{3}{8}$
$x \log 3 = \log \left(\frac{3}{8}\right) = \log \left(3\right) - \log \left(8\right)$
$x = \left(\log \frac{3}{\log} \left(3\right)\right) - \left(\log \frac{8}{\log} \left(3\right)\right) = 1 - 1.89 = - 0.89$