# exponential equation different base and 2x

I am "absolute beginner" and I am working on this equation:

$$\frac{1+5^{(2x+1)}}{4} = 5^x$$

It's all day that I am working on it but I don't seem to be able to find a solution. This is my last test:

$$1/4 + 5^{2x} + 5/4 = 5^x$$

$$1 + 5^{2x} = 5^x$$

Wondering if it can be correct until now? Here I am bit stuck. I thought to use logaritms, but I can't find a nice solution.

Any hints appreciated, I would really like to learn it and not looking just for the solution. (I am trying to read as much as I can similar answers).

thanks

Set $\displaystyle3^x=a\implies3^{2x+1}=3(3^x)^2=3a^2$ to find $$1+3a^2=4a\iff3a^2-4a+1=0$$

$\implies a=1=3^0,\dfrac13=3^{-1}$

Now for $\displaystyle c\ne0,\pm1;$ $c^x=c^y\implies x=y$

• sorry, I am not able to understand your answer :/ – user181528 Oct 8 '14 at 6:28

Wondering if it can be correct until now?

Well, no. Note that $$\frac{1+3^{2x+1}}{4}=\frac 14+\frac{3^{2x+1}}{4}=\frac 14+\frac{3}{4}\times 3^{2x}.$$ So, we have $$\frac 14+\frac{3}{4}\times (3^x)^2=3^x$$ Then, set $t=3^x$ to get $$\frac 14+\frac 34t^2=t$$ which should be easy to do with.

• agh so the first error was just on the exponents of the fraction! That I've put a sum instead of multiplication right? – user181528 Oct 7 '14 at 15:28
• @jsabina: Yes, I guess so. – mathlove Oct 7 '14 at 15:33