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Are you asking for theory or practice?
For practical applications, it is quite easy to learn and/or calculate the log table. Start with learning that log 2 = 0.3, and of course, log 3 is approx. 0.5. Since log 4 = 2(log 2) = 0.6; log 6 = log 2 + log 3 = 0.8. So, log 5 is between log 4 and log 6, or about 0.7, and so forth.
Such methods for estimation have largely been lost with generations that have grown up with handheld calculators, but they can still be useful to know.
In your example, log (5^3.4) = 3.4 log 5 = 3.4X0.7 = 2.38
Antilog(2.38) = 100X (antilog 0.38)
Antilog (0.38) is between antilog (0.3), i.e., 2, and antilog (0.4). At this point, there are different ways to solve for antilog (0.4)
For example:
Antilog 0.4 = 0.8/2 = (log 6)/2 = 6^1/2
We know that 2.5^2 = 6.25 (i.e., 25^2 = 625)
Thus, you know that antilog 0.4 is less than 2.5, but close to it
So, an estimate for the antilog of 0.38 is between 2.0 and 2.5 and closer to 2.5
If you chose 2.4, your scratch pad calculation of 5^3.4 comes out to be 240.
My handheld calculator shows the answer as: 238
John