1. ## Exponential equation help

Need help with problem. I think it is a characteristic equation where u let y= something

5 2x-1 + 5x=10

2. ## Re: Exponential equation help

Is the equation:

$\displaystyle 5^{2x-1}+5^x=10$ ?

3. ## Re: Exponential equation help

Yes that is the equation. Thanks and sorry for the mistype

4. ## Re: Exponential equation help

To express the equation in plain text, you could write:

5^(2x - 1) + 5^x = 10

To use $\displaystyle \LaTeX$ wrap the following code:

5^{2x-1}+5^x=10

In TEX tags, which you can generate using the $\displaystyle \Sigma$ button on the toolbar. This will give you:

$\displaystyle 5^{2x-1}+5^x=10$

For this equation, I would use a bit of trial and error. Let's let $\displaystyle x=0$:

$\displaystyle 5^{2(0)-1}+5^0=\frac{1}{5}+1=1.2<10$

So, we see $\displaystyle 0<x$. Next try $\displaystyle x=1$...what do you find?

5. ## Re: Exponential equation help

Nothing. Couldn't figure it out. I'm still trying.

6. ## Re: Exponential equation help

I think I got it. X=1

7. ## Re: Exponential equation help

Originally Posted by Nfalconer
I think I got it. X=1
Yes, when x = 1, we get:

5 + 5 = 10

and this is true.

8. ## Re: Exponential equation help

What we should really do here, and I apologize for being thick, is take the original equation and multiply through by 5 to get:

$\displaystyle 5^{2x}+5\cdot5^x=50$

Write as quadratic in $\displaystyle 5^x$ in standard form:

$\displaystyle \left(5^x\right)^2+5\cdot5^x-50=0$

Factor:

$\displaystyle \left(5^x+10\right)\left(5^x-5\right)=0$

As $\displaystyle 0<5^x$ for all real $\displaystyle x$, we are left with:

$\displaystyle 5^x-5=0$

$\displaystyle 5^x=5^1\implies x=1$

9. ## Re: Exponential equation help

Originally Posted by MarkFL

$\displaystyle \left(5^x\right)^2+5\cdot5^x-50=0$
Or, the original poster referenced substitution. Let y = $\displaystyle \ 5^x$.
Then factor $\displaystyle \ y^2 - 5y - 50 = 0$.

Solve that for y. And then substitute back for x, noting the allowable real value of x.

10. ## Re: Exponential equation help

I don't understand the substitution way of solving the equation. Could you elaborate?

11. ## Re: Exponential equation help

Originally Posted by Nfalconer
I don't understand the substitution way of solving the equation. Could you elaborate?
$5^{2x-1}+5^x=10$ let $y=5^x$ then we get $(5^x)^2(5^{-1})+5^x-10=0$ so that
\begin{align*}(5^{-1})y^{2}+y-10&=0 \\y^2+5y-50&=0\\(y+10)(y-5)&=0\\(y+10)~&\wedge~(y-5)=0\\y=-10 ~&\wedge~y=5\end{align*}

From that we get $5^x=-10~\wedge~5^x=5$