Question about Casio Office Equipment & Supplies

# How to calculate antilog on a KS-570MS?

Karson KS-570MS calculator

Posted by Anonymous on

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Posted on Jan 02, 2017

I reccently had the same problem, but I have a Ti-84 Plus,and I am disgusted by how poorly this matter is covered by Texas Intruments. Esp. since the solution is rather simple.

The antilog key is [2nd] [log] or the 10^x. On your screen it should apear as 10^( and then you just input the value that you want to find the antilog of.

When used to solve pH problems remember to input the pH value as a negative number, since pH= -log[H+] then [H+]= antilog(-pH).

Hope that helped))))) ^^

Posted on Nov 25, 2008

pH is minus (log to base 10) of the hydrogen ion activity of an aqueous solution, or (log to base 10) of (1/hydrogen ion activity)

To get the inverse log, i.e the hydrogen ion activity corresponding to a specified pH, simply enter the pH value and press

2nd
LOG
1/x

Posted on Jan 01, 2009

I'm not specifically familiar with the TI83 or TI84 but I've used a lot of TI calculators in my time, so I'll give it a try. If your trying to find the antilog of a number in base 10 enter the number and hit the (10 to the X) button. If you're trying to find the antilog of a number in in base e (natural log), enter the number and hit the (e to the X) button.

Posted on May 29, 2009

Anti log will be shown as 10^x (10 raised to power x) For natural log it's e^x I think it's shift > log

Posted on Jun 03, 2009

Hello,

Let y=10^(x) 10 to the power of x
Take the log of both tems of the equality. You get
log(y)=log[10^(x)] where I used square brackets for clarity. But from the general properties of logarithms

log(b^(a)) = a*log(b)
Applied to our expression above
log(10^x)=x*log10
But since we are using log as log in base 10, log_10(10)=1
so
log(y)=x
We thus have two equivalent relations
y=10^x <----> x=log(y) The double arrow stands for equivalence.

If y is the log of x, then x is the antilog of y

Your question: With log_10 standing for logarith in base 10
pH=-log_10(c) c= concentration
Then log_10(c)=-pH the equivalence above translates as
log_10(c)=-pH is equivalent to c=10^(-pH)

The concentartion corresponding to a pH of 7.41 is
c=10^(-7.41)=3.89x10^(-8)

You enter it as follows
10[^]7.41[(-)][ENTER/=]

Hope it helps

Posted on Sep 07, 2009

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