# Indices and Algebra

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

• Let $u=5^x$ and then solve the quadratic. – Donald Splutterwit Nov 17 '17 at 0:03
• Do you mean \begin{eqnarray*} 2 \times 5^{x+1}=1+\frac{3}{5^x} \end{eqnarray*} now ? – Donald Splutterwit Nov 17 '17 at 0:12
• yes that was how it was asked – Corradi Nov 17 '17 at 0:14

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 $$5^x=1$$ $$\Leftrightarrow x=0$$