Find all the possible value of x in this equation \begin{eqnarray*} 2\times 5^{x+1}=1+\frac{3}{5^x} \end{eqnarray*} Thanks in abundance
I don't know how to use iteratiom to kill it. Pls help me out
0
$\begingroup$
$\endgroup$
-
$\begingroup$ Let $u=5^x$ and then solve the quadratic. $\endgroup$ – Donald Splutterwit Nov 17 '17 at 0:03
-
$\begingroup$ Do you mean \begin{eqnarray*} 2 \times 5^{x+1}=1+\frac{3}{5^x} \end{eqnarray*} now ? $\endgroup$ – Donald Splutterwit Nov 17 '17 at 0:12
-
$\begingroup$ yes that was how it was asked $\endgroup$ – Corradi Nov 17 '17 at 0:14
2
$\begingroup$
$\endgroup$
We have to solve the equation $$ 2\cdot (5^x+1)=1+\frac{3}{5^x} $$ Let's multiply its both sides by $5^x$ to get rid of the unpleasant fraction: $$ 2\cdot 5^x\cdot(5^x+1)=5^x+3 $$ \begin{align} &\Leftrightarrow 2\cdot(5^x)^2+2\cdot 5^x=5^x+3\\ &\Leftrightarrow 2\cdot(5^x)^2+5^x-3=0 \end{align} Put $y=5^x$. The latter equation can be expressed now in the form $$ 2y^2+y-3=0 $$ And we solve it in $y$: $$ \Delta=25 $$ $$ y_1=\frac{-1-\sqrt{\Delta}}{4}=-\frac{3}{2} $$ $$ y_2=\frac{-1+\sqrt{\Delta}}{4}=1 $$ Now it is the time to look back how we have defined $y$, namely $y=5^x$. This means that $y$ cannot be negative and in result, only the second root is possible. Hence $$ y=1 $$ which gives $$ 5^x=1 $$ $$ \Leftrightarrow x=0 $$
