# How do you solve 5^(2x-1) -2 =5^(x-1/2)?

Jun 21, 2016

$x = \frac{1}{2} {\log}_{5} \left(20\right) .$

#### Explanation:

Let $\left(2 x - 1\right) = t .$

Observe that $R H S$ of the Eqn. is ${5}^{\frac{2 x - 1}{2}} .$

Hence, the given eqn. becomes, ${5}^{t} - 2 = {5}^{\frac{t}{2}}$.

Next, let ${5}^{\frac{t}{2}} = y$, so that, ${5}^{t} = {\left\{{5}^{\frac{t}{2}}\right\}}^{2} = {y}^{2.}$

Now the eqn. is, ${y}^{2} - 2 = y ,$ or, ${y}^{2} - y - 2 = 0.$

$\therefore {y}^{2} - 2 y + y - 2 = 0.$
$\therefore y \left(y - 2\right) + 1 \left(y - 2\right) = 0.$
$\therefore \left(y - 2\right) \left(y + 1\right) = 0.$
$\therefore y = 2 , y = - 1 ,$ but, $y$ being ${5}^{\frac{t}{2}}$ can not be $- v e$, so, $y \ne - 1$.
$\therefore y = 2 \Rightarrow {5}^{\frac{t}{2}} = 2 \Rightarrow {5}^{t} = {\left\{{5}^{\frac{t}{2}}\right\}}^{2} = {2}^{2} = 4.$
$\therefore t = {\log}_{5} \left(4\right)$
$\therefore 2 x - 1 = {\log}_{5} \left(4\right) .$
$\therefore 2 x = 1 + {\log}_{5} \left(4\right) , = {\log}_{5} \left(5\right) + {\log}_{5} \left(4\right) = {\log}_{5} \left(20\right) .$
$\therefore x = \frac{1}{2} {\log}_{5} \left(20\right) .$