Anti Logarithm Formula Is there any formula to calculate anti logarithm just using simple calculator.. I already know how to calculate logarithm digit by digit exactly just like this What is the best way to calculate log without a calculator? But how to calculate the antilog like antilog(0.456)=10^0.456=2.857590 using simple calculator..
 A: I would use Taylor polynomial. In case of function $f(x)=10^x,$ you have
$$f(x)=1+x\ln10+\frac{(x\ln10)^2}{2!}+\frac{(x\ln10)^3}{3!}+\frac{(x\ln10)^4}{4!}+\cdots$$
For a decent approximation, it is usually enough to use first 6 or 7 elements of the series. In this particular case
$$10^{0.456}\approx1+.456*2.3026+\frac{(.456*2.3026)^2}{2}+\frac{(.456*2.3026)^3}{6}+\frac{(.456*2.3026)^4}{24}+\frac{(.456*2.3026)^5}{120}+\frac{(.456*2.3026)^6}{720}=2.8572\ldots$$
Another idea that comes to mind is using Newton's method.
[edit]
Calculating $a^b$ using Newton's method would look like this:
Step 1: Choose an initial value (your best guess) $x_0$
Step 2: Calculate $x_{n+1}=x_n-(\log_a x_0-b)x_0\ln a$
Step 3: Repeat step 2 until satisfied with the approximation.
Your example:
$$x_0=3$$
$$x_1=3-(\log 3-0.456)3\ln 10=2.8541$$
$$x_2=2.8541-(\log 2.8541-0.456)2.8541\ln 10=2.85759$$
This method requires fewer steps to achieve the same precision level but the steps are more complex, especially if you are calculating the logarithms by hand (or using a simple calculator).
A: I realize I'm a little late to the party, but if you need a simple way to calculate a rough approximation of antilog base 10, use the following procedure:

*

*Separate the integer part from the fractional part.  So 1.3979 is split into 1 and 0.3979.


*Use the fractional part on this formula:
5.5x^3+  x^2 + 2.5x + 1
So in this example, 5.5(0.3979^3) + 0.3979^2 + 2.5(0.3979) + 1
which equals: 2.499 or 2.5 with a little rounding.


*Now write the answer and then E and then the integer part of the original number.
So we get 2.5E1 which is short for 2.5x10^1 or 25.
If you aren't looking for exact values and just need to be in the ballpark, this formula will usually do.  If you are making some kind of graphics program where speed is more important than accuracy, then this is what you want.
