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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.

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up vote 10 down vote accepted

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

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 $5^a \cdot 5^b = 5^{a+b}$

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Could you elaborate on this further and provide a bit more background? –  user493941 Mar 20 '13 at 18:56
    
@user493941 ok i tried to explain it in an easy way, does it help? –  Dominic Michaelis Mar 20 '13 at 19:03
    
Wow! Great explanation! –  user493941 Mar 20 '13 at 19:10
    
@downvoter why? –  Dominic Michaelis Mar 20 '13 at 23:07
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the injectivity of the exponential function $f(x) = 5^x$

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

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 $2x+1-2=0$.

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Why didn't you just take the $log_5$ of the original formula to get $2x+1=2$ and avoid the division? –  Thomas Andrews Mar 20 '13 at 19:20
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I think $$5^{(2x+1)} = 5^2$$

$$\implies$$

$$\log_5 5^{(2x+1)} = \log_5 5^2$$

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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.

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