# Why does $10^{\log_{10}2} = 2$? [closed]

Why does $10^{\log_{10}2} = 2$ ?

I tried to draw a graph but that didn't help

• The definition of $\log$ is as follows: $\log_a b=c\iff a^c=b$. If $c=\log_{10} 2$, then $10^c=2$, Q.E.D. Commented Feb 7, 2018 at 6:00
• How did you draw a graph. Commented Feb 7, 2018 at 6:01
• Because $\,f(f^{-1}(x)) = x\,$ by definition. What's your definition of $\,\log_{10}(\,\cdot\,)\,$?
– dxiv
Commented Feb 7, 2018 at 6:01
• @KingTut Tut use a computer program. Commented Feb 7, 2018 at 6:03
• The value of $\log_{10}2$ is, by definition, the number $x$ such that $10^x=2$. So what happens when you plug it in for $x$? It makes the equation true.
– anon
Commented Feb 7, 2018 at 6:03

The definition of log base $b$ of $x$, or $log_{b}(x)$, is the solution $y$ such that $b^y = x$. Therefore, this is true given the definition.

If $$\log_a c = b$$ then this means that $$a^b = c.$$ The latter is just simply another way of writing the former logarithm. Therefore, if we consider the case where $a = 10$ and $c = 2$ then we have$$\log_{10} 2 = b$$ which means that $$10^b = 2$$ and thus we arrive at the conclusion that $$10^{\log_{10}2} = 2.$$ And, because $10 = a$ and $2 = c$, then we arrive at a more general conclusion: $$a^{\log_a c} = c.$$ We can now put, like how we would similarly put in division, that the base $a$ and $\log_{a}$ cancel out to yeild c.

Therefore in your case, the base $10$ and $\log_{10}$ cancel out to yeild $2$.

$\log_a x=log_a x$

Due to definition:

⇒ $x=a^{log_a x}$

$log _{10}2=log_{10}2$ ⇒ $2=10^{log_{10}2}$