How do I solve $2\left(5^x\right)+5^{-x}=3$ for $x$? I understand logs and can solve simpler equations like
$$
7^x=3 \\
\log7^x=\log3\\
x = \frac{\log3}{\log7}
$$
But the addition in this particular expression really throws me off and I have no idea how to even approach it:
$$
2\left(5^x\right)+5^{-x}=3
$$
If it was a multiplication, no problem, but I have never seen problems with addition like this before.
I have tried rearranging and factorising, e.g. $5^{-x}\left(2\left(5^{2x}\right)+1\right)=3$, but can someone show me what I need to do to solve problems of this form?
 A: Hint.
You can use the substitution $y=5^x$.
Then your equation becomes
$$2y+\frac 1y=3$$
which is a simple quadratic equation.
A: I remember the first time I saw a problem like this. Totally stumped me. You need a new sort of "idea" to solve something like this which your teacher probably hasn't taught you. 
There really isn't an easy way to solve this equation as is, so we'd like to simplify it somehow to make it look like something we are more familiar with, so we use a technique called "substitution".
You let some term in this equation be equal to some new variable so that the equation will look more like something we are familiar with. 
What might that be in this case? Try letting $y=5^x$. Do you see how this simplifies your equation? Can you solve for y from there? 
Don't forget to solve for x (your final answer) once you solve for y though! 
The idea of substitution is an important one that if you go on in mathematics towards calculus 2 you will be seeing a lot! 
A: you can put $y=5^x$
so $2y+\frac{1}{y}=3$ $\rightarrow$$ 2y^2+1-3y=0$
$\triangle =9-8=1$
so , $Y=1$ OR $y=\frac{1}{2}$
finally;
$y=5^x=1$$\rightarrow$ $ x=0$
or
$y=5^x=\frac{1}{2}$$\rightarrow$ $ x=\frac{-ln(2)}{ln(5)}$
