# Using log tables for exponential solutions

I understand how to use a log table to solve something such as $$\log(0.00000000453)$$ where we would put $$(0.000000453)$$ into scientific notation, $$4.53 \times 10^{-9}$$. Then we can use the log table to find the mantissa of the log, which is $$0.6561$$, and use the characteristic, $$-9$$ to add together and get $$0.6561+(-9)=-8.3439$$ and $$\log (0.00000000453)=-8.3439$$.

However, if I am given $$1.64^{28}$$, how would I use the log table? I can use log properties and get $$28 \times \log 1.64 = 28 \times 0.2148$$ (value from log table). But this gives me $$6.0144$$ which is not $$1.64^{28}=1036639.481$$. How do I take my log table calculation and get back to the exponential answer?

Since $$\log 1.64^{28} = 6.0144$$, then $$1.64^{28} = 10^{6.0144} = 10^6 \times 10^{0.0144}.$$ Find in the tables a certain number $$x$$ such that $$\log x \simeq 0.0144$$, and then let $$1.64^{28} = x \times 10^6$$.

• But all I want is to solve $1.64^{28}$, no log. – K Math Oct 16 '18 at 20:14
• What do you mean by "solve" a number? – Hugo Oct 16 '18 at 20:15
• Find the equivalence to. I used a calculator to find $1.64^{28}=1036639.481$ but I want to use log tables to figure this out. So far I've used them to find $log 1.64^{28}$ not just $1.64^{28}$. – K Math Oct 16 '18 at 20:16
• Edited, hope it's clearer now. – Hugo Oct 16 '18 at 20:23

$$6.0144$$ is the log of $$1.64^{28}$$.
Then you have to look in the tables in the reverse: as for which number has a log of $$0.0144$$ (in case interpolating) and then add $$\times 10^6$$. The all apart of course from truncation errors (the actual $$log 1.64^{28}=6.0156277...$$)

• So once I've found the number that has log $0.0144$ and add $10^6$ I still do not get an answer equivalent to $1.64^{28}$. Am I misunderstanding? – K Math Oct 16 '18 at 20:20
• At the time I was a student we really had just the log tables, and we were aware of possible truncation errors: if we were looking for $0.0144$ and knew it was truncated we had also to give a glance to what $0.01445$ and or $0.01435$ was. – G Cab Oct 16 '18 at 20:38
• @KMath, in particular, taking only 3 significative digits and multiply by $28$ will of course give a large error. – G Cab Oct 16 '18 at 20:44