Logariths

[webplodder]

I know a little about logarithms, for example, I know that in the case of 2^3 = 8, 3 is the logarithm of 8 to base 2, or 3 = log2, but how to fractional logs work, for example, 10^1.5? I'm not mathematically literate so a simple explanation would be nice.

 

Thanks.

[Country Boy]

10^1.5= 10^(3/2)= sqrt(1000). According to the calculator included with Windows, that is 31.6227.... .Actually with the Windows calculator it is easier to do it directly using the "x^y" key. Enter 10, then click "x^y", then enter 1.5 and =.

 

Notice that I did 10^(3/2)= sqrt(1000) by doing 10^3 first, then (1/2) power= square root. You could also do it the other way: take the square root of 10, then the third power. Of course, if you are doing it "by hand", that second way is harder.

[Dave]

I know a little about logarithms, for example, I know that in the case of 2^3 = 8, 3 is the logarithm of 8 to base 2, or 3 = log2, but how to fractional logs work, for example, 10^1.5? I'm not mathematically literate so a simple explanation would be nice.

 

Thanks.

 

There's very little to learn past this point

 

Essentially, we just define [math]a=\log_b c[/math] as the number which satisfies the equation [imath]b^a = c[/imath]. That's all it is - we can derive everything else about logarithms from this point.

 

Perhaps if you identified precisely what's troubling you I could help more?

[Country Boy]

I responded telling how you can calculate 10^1.5 (it's about 31.6). If your question was how to convert the "exponential" equation, 10^1.5= 31.6, to a "logarithmic" equation, the answer is exactly what you did with 2^3= 8. The base of the exponent, here 10, becomes the base of the logarithm, the "result", here 31.6, becomes the argument of the logarithm, and the argument of the exponential becomes the "result" of the logarithmic equation:

10^1.5= 31.6... becomes log_10(31.6...)= 1.5.

 

As dave said, the exponential equation a^x= y becomes log_a(y)= x.

Since "exponential" and "logarithm" (to the same base) are INVERSE functions, changing from one to the other just swaps "argument" and "result" (x and y).