Using log table to solve a division problem

Question

Given $187,000,000,000 \div 0.00000000453$, we can put our givens in scientific notation, i.e. $1.87 \times 10^{11} \div 4.53 \times 10^{-9}$. Now we can take the log and use the log table: $11.2718 \div -8.3439$ (after using log table). Now I'm unsure how to continue. If I compute $11.2718 \div -8.3439$ I get $-1.350903055$. Our original problem has an answer of $4.1280 \times 10^{19}$. Wat am I missing?

Answer 1

Instead of dividing 11.2718 by −8.3439, you should subtract the latter from the former.

Answer 2

Notice that $$\log\left(\frac{a}{b}\right) = \log a - \log b$$ for $\frac{a}{b}>0$.

Answer 3

To divide two number, subtract their logarithms first to get the logarithm of the quotient:

$$11.2718-(-8.3439)=20.6157$$ Now look up the mantissa $0.6157$ in your table, you should get $4.1276$. So the final answer is $4.1276\cdot10^{20}$.

Answer 4

Use the fact that

$$\log\left(\frac{1.87 \times 10^{11}}{4.53 \times 10^{-9}}\right) \;\; = \;\; \log\left(1.87 \times 10^{11}\right) \; - \; \log\left(4.53 \times 10^{-9}\right) $$

$$ = \;\; \left(\log 1.87 \; + \; \log 10^{11}\right) \; - \; \left(\log 4.53 \; - \; \log 10^{-9}\right) $$

$$ = \;\; (\log 1.87 \; + \; 11) \; - \; (\log 4.53 \; - \; 9) $$

Here's how the work was arranged when I was in school. After expressing the input values in scientific notation, you first write the following. (The stuff I wrote above is for your benefit, not something we would actually write.)

$$\begin{array}{cccc} & \log 1.87 & + & 11 \\ (\text{minus})& \log 4.53 & - & 9\\ \end{array}$$

Then we would look up the logarithms in the back of the textbook and write

$$\begin{array}{cccc} & 0.2718 & + & 11 \\ (\text{minus})& 0.6561 & - & 9\\ \end{array}$$

Since the "mantissa number" being subtracted (i.e. $0.6561$) is larger than the "mantissa number" it's being subtracted from (i.e. $0.2718),$ we make an adjustment to the top number by adding and subtracting $1$ to get

$$\begin{array}{cccc} & 1.2718 & + & 10 \\ (\text{minus})& 0.6561 & - & 9\\ \end{array}$$

Now perform the subtraction, remembering that subtacting $-9$ is the same as adding $9.$

$$ 0.6157 \; + \; 19 $$

Now we would look up the antilog of $0.6157$ and multiply it by $10^{19}$. That is, we use the fact that $10^{0.6157 + 19} = 10^{0.6157} \times 10^{19},$ where $10^{0.6157}$ is the antilog of $0.6157.$ Doing this gives

$$ 4.128 \; \times \; 10^{19}$$