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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?

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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 $1.64^{28} = x \times 10^6$.

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  • $\begingroup$ But all I want is to solve $1.64^{28}$, no log. $\endgroup$ – K Math Oct 16 '18 at 20:14
  • $\begingroup$ What do you mean by "solve" a number? $\endgroup$ – Hugo Oct 16 '18 at 20:15
  • $\begingroup$ 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 $1.64^{28}$. $\endgroup$ – K Math Oct 16 '18 at 20:16
  • $\begingroup$ Edited, hope it's clearer now. $\endgroup$ – Hugo Oct 16 '18 at 20:23
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$6.0144$ 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 $log 1.64^{28}=6.0156277...$)

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  • $\begingroup$ 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? $\endgroup$ – K Math Oct 16 '18 at 20:20
  • $\begingroup$ 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. $\endgroup$ – G Cab Oct 16 '18 at 20:38
  • $\begingroup$ @KMath, in particular, taking only 3 significative digits and multiply by $28$ will of course give a large error. $\endgroup$ – G Cab Oct 16 '18 at 20:44

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