A quick Logarithm question Did I do this problem right?
$\log y= 10 + 2(\log x)$
$y= 10^{10} + x^2$?
Edit: How does $10^{\log (x^2)}= x^2$? I can't seem to find an explanation for this. All the websites that I found already assumes that it equals $x^2$. Sorry for the trouble.

I'm a bit lost :(. 
 A: You had a good idea by raising $10$ to the power of each side.
Note, however, that $a = b+c$ implies that $10^a = 10^{b+c} = 10^b\times 10^c$
So,
$\log_{10} y = 10 + \log_{10}(x^2)$
$10^{\log_{10}y} = 10^{10+\log_{10}(x^2)} = 10^{10}\times 10^{\log_{10}(x^2)}=\dots$
A: just remember that $a^{\log_a x}=a$ & $\log_a x=y$ then $a^y=x$. I think you have wrighten wrong, it has to be $y= 10^{10}\times x^2$.
or $\log a-\log b=\log\dfrac{a}{b}$ $$\log y= 10 + 2(\log x)=10+\log x^2$$
$\log y-\log x^2=10$ thus $\log \dfrac{y}{x^2}=10$ then $\dfrac{y}{x^2}= 10^{10}$ hence $y= 10^{10}\times x^2$.
A: Here is why $10^{\log \left(x^2\right)}= x^2$. It's almost exactly the definition of logarithm.

Definition of logarithm: $\log_a b=c\iff a^c=b$

In this case: $\log \left(x^2\right)=c\iff 10^c=x^2\ \ \ (1)$
Now remember: $a=b\iff 10^a=10^b$ 
(it's because $f(x)=10^x$ is a strictly increasing function).
Therefore: $\log\left(x^2\right)=c\iff 10^{\log\left(x^2\right)}=10^c$, which by $(1)$ equals $x^2$.

So to complete the problem:
$\log(y)=10+2\log x\iff 10^{10+2\log x}=y$ (by definition).
$10^{10+2\log x}=10^{10}\cdot 10^{2\log x}$ by exponent property: $a^{b+c}=a^b\cdot a^c$.
$10^{10}\cdot 10^{2\log x}=10^{10}\cdot 10^{\log \left(x^2\right)}$ by logarithm property: $b\log a=\log a^b$.
$10^{10}\cdot 10^{\log \left(x^2\right)}=10^{10}\cdot x^2$ (see beginning of the answer).
