Function application (word problem)

Question

The problem:

My work so far:

$3=log(\frac{A}{A_0})$--->$10^3=\frac{A}{A_0}$

$\frac{A}{A_0}=1000$

(Am I done there?)

Plugging it in:

$M=log(\frac{1900000}{1000})$

$10^M = \frac{1900000}{1000}$

$M=3.278753601$

I know this is wrong because for it to be 10 times as strong it would have to be a 4.0

EDIT: I realize that $A_0$ can't 1000 because $A_0$ Is supposed to be the smallest measurable quake.

Answer 1

You have it correct that $\frac{A}{A_0}=1000$ for a quake of magnitude $3.0$. Now you want to find the Richter value of a quake $1900$ times as strong as this, so you take $$\log{(1900\cdot1000)} \approx 6.28$$

Your division by $1000$ is unnecessary. It makes it as though a magnitude $3.0$ earthquake is the basis of the Richter scale, making it the new $0$ point.

Answer 2

You have $$ \frac{A}{A_0}=1000\quad\Rightarrow\quad A_0=\frac{A}{1000} $$ then $A'=1900A$. $$ \begin{align} M'&=\log_{10}\left(\frac{A'}{A_0}\right)\\ &=\log_{10}\left(\frac{1900A}{\frac{A}{1000}}\right)\\ &=\log_{10}(1900000)\\ &=\log_{10}(1.9\times10^6)\\ &=\log_{10}(1.9)+\log_{10}(10^6)\\ &\approx6.27875 \end{align} $$