In Logarithms and Antilogarithms,calculation of Logarithms seems easy after knowing the properties,calculation of Antilogs seems entirely difficult without using the Log-Antilog Table.I've tried several conventional methods:
Split it into smaller terms,take approximation of ranges,etc. but it's too tedious and time-consuming for me.Is there perhaps a different method/faster method to calculate the Antilog of a number without using the Log-Antilog Table?
Can anyone please help me on this topic?
PS:I'm not talking about natural logs,just the standard base 10 logs-antilogs.
$\endgroup$ 91 Answer
$\begingroup$Write the logarithm as$$\log_{10}(y)=\lfloor \log_{10}(y)\rfloor +a$$ So$$y = 10^{\lfloor \log_{10}(y)\rfloor } \times 10^a=10^{\lfloor \log_{10}(y)\rfloor } \times e^{a \log_e(10)}$$ Now, with $k=a \log(10)$, use one of the approximations of $e^k$ such as$$e^k\sim\frac{2+k}{2-k} \qquad \text{or}\qquad e^k\sim\frac{12+6k+k^2}{12-6k+k^2} \qquad \text{or}\qquad e^k\sim \frac{120+60k+12 k^2+k^3}{120-60k+12 k^2-k^3}$$
Example
As asked in comments,$$\log_{10}(y)=5.7=5+0.7 \implies k=0.7\log_e(10)$$ Using the last approximation$$ e^k\sim \frac{120+60k+12 k^2+k^3}{120-60k+12 k^2-k^3}=5.01343\implies y=5.01343\times 10^5$$
In fact, since $a >0.5$, it is better to work with ceiling$$\log_{10}(y)=5.7=6-0.3 \implies k=-0.3\log_e(10)$$ Using the last approximation$$ e^k\sim \frac{120+60k+12 k^2+k^3}{120-60k+12 k^2-k^3}=0.501187\implies y=0.501187\times 10^6=5.01187\times 10^5$$
The exact value is $y=5.01187\times 10^5$
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