SOLUTION: Solve antilog(-3.5048)
Algebra
->
Numeric Fractions Calculators, Lesson and Practice
-> SOLUTION: Solve antilog(-3.5048)
Log On
Algebra: Numeric Fractions
Section
Solvers
Solvers
Lessons
Lessons
Answers archive
Answers
Click here to see ALL problems on Numeric Fractions
Question 1049708
:
Solve antilog(-3.5048)
Answer by
Theo(13342)
(
Show Source
):
You can
put this solution on YOUR website!
the anti-log of a number is the number that you take the log of in order to get that number.
so, if your number is -3.5048, then the anti-log of -3.5048 is the number that you take the log of in order to get -3.5048.
algebraically this is shown as log(x) = y.
unless otherwise stated, the base of the log is always assumed to be 10.
in your problem this would be log(x) = -3.5048.
x is called the anti-log of -3.5048.
you need to solve for x.
the basic definition of logs states:
log(x) = y if and only if 10^y = x.
in your problem, y is equal to -3.5048, so the definition becomes:
log(x) = -3.5048 if and only if 10^-3.5048 = x.
you can use your calculator to solve for x to get x = (3.127519311 * 10^-4).
if this is correct, then 10^(3.127519311 * 10^-4) will be equal to -3.5048.
use your calculator to see that it does.
note that 3.127519311 * 10^-4 is the same as .0003127519311.
you can enter it either way.
here's an online calculator you can use if you don't have one.
http://www.alcula.com/calculators/scientific-calculator/
here's a picture of my calculations on this calculator.
you simply enter 10^-3.5048 and hit the enter key and you will get 0.000312752 as shown on the calculator display.
the place where you make your entry is the space right above the keys.
the display is on top of that.
you then enter log(0.000312752) and the calculator will tell you that the answer is 3.504799904.
the confirmation is not right on because there is rounding going on.
here's a tutorial on the basic relationship between logs and exponents.
http://www.purplemath.com/modules/logs3.htm
i'm assuming you can use a calculator to find the anti-log.
if you have to use a log table instead, then let me know and i'll take you through that. after i refresh my memory on how.