What is the property that allows $5^{2x+1} = 5^2$ to become $2x+1 = 2$? What is the property that allows $5^{2x+1} = 5^2$ to become $2x+1 = 2$? We just covered this in class, but the teacher didn't explain why we're allowed to do it.
 A: the injectivity of the exponential function $f(x) = 5^x$
A: Its a really good question, the thing is that $5^{x}$ is injective, so that it is only possible that $5^x = 5^y$ when $x=y$. 
You know that $a^b \cdot a^c= a^{b+c}$, so when you divide by $5^y $ we have
$$5^{x-y}=1$$
when $x-y\neq 0$ this won't be 1.
I guess it is easiest to unterstand when you use that strict monotonicty implies injectivity.
Because when $x-y>0$ we will multiply a bit of $5$  to it and as bigger $x-y$ is the bigger the number is we multiply. When $x-y<0$ we divide trough a bigger number as $|x-y|$ gets bigger.

Here is a plot of the function in the interval $[-1,1]$, in fact it is sufficient to prove that the function is monotone here as we know that we can use for any other number a sum of numbers in this intervall and the above mentioned rule $5^a \cdot 5^b = 5^{a+b}$ 
A: I think $$5^{(2x+1)} = 5^2$$
$$\implies$$
$$\log_5 5^{(2x+1)} = \log_5 5^2$$ 
A: $5^{(2x+1)} = 5^2$
Multiplying by $1/5^2$ om both sides we get,
$\frac{5^{(2x+1)}}{5^2} = 1$
$\Rightarrow 5^{(2x+1)-2} = 1$
Taking log to the base 5 on both sides we get $2x+1-2=0$.
A: Everyone seems to have mentioned injectivity, but, I suppose, if you're asking a question like this then you may well not know what it means for a function to be injective.
A function $f: X \to Y$ is a mapping which assigns to each element of the set $X$ a unique element in the set $Y$.
We say that $f$ is injective if, $\forall f(a),f(b) \in Y$ we have that $f(a) = f(b) \implies a = b$. (Of course, this only makes sense if $a, b \in X$. Now you can consider this with the function $a^x, a \in \mathbb{R}$ to see how things work.

Alternatively, you could say that
$5^{2x+1} = 5^2 \implies 5^{2x - 1} = 5^0 = 1 \implies 2x - 1 = 0 \implies 2x + 1 = 2$
if you're prepared to believe that the only number which when you raise $5$ to the power of said number gives you $1$, is $0$. However, the justification for this is the above notion of injectivity, so this kind of approach is washing over the point of the deduction.
