A quick Logarithm question

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.

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I'm a bit lost :(.

Answer 1

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 $y= 10^{10}\times x^2$.

Answer 2

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,

$\log_{10} y = 10 + \log_{10}(x^2)$

$10^{\log_{10}y} = 10^{10+\log_{10}(x^2)} = 10^{10}\times 10^{\log_{10}(x^2)}=\dots$

Answer 3

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 $x^2$.

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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).