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In the TI-98 Titanium calculator, the default for log is 10, how do I either change it to base 2 or how to do I enter values to make it base to. eg: log(9/14) = -.637

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To compute a log to any base, enter it as log(x, base). You can either select log from the catalog or simply type it in.

Posted on Nov 15, 2010


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Change base log

The TI 86 has two logarithmic functions: natural logarithm (ln) and common (decimal) logarithms (log). If you need the logarithm in any other base than e or 10 you need to use one of the two equivalent expressions
log_b(x) =ln(x)/ln(b) =log(x)/log(b)
Here b is the value of the base of the logarithm and x is the argument (the value whose logarithm you are seeking). Of course the argument x must be a positive number.
Note: On the TI 86 the log function can calculate the logarithm of a complex number, according to the manual.

Sep 22, 2013 | Texas Instruments TI-86 Calculator

1 Answer

Where can I find the log values in TI 89

For natural logarithms, you can use the ln function (2ND shift of the X key). For example, the natural logarithm of 3 is 2ND [LN] 3 ) ENTER.

For logarithms to other bases, use the log function (diamond shift of the 7 key). For example, the base-2 logarithm of 16 is diamond 7 1 6 , 2 ) ENTER. If the base is 10 (common logarithms), you can omit the comma and the second argument.

Mar 09, 2011 | Texas Instruments TI-89 Calculator

1 Answer

How do you enter the log base on the TI 89 Titanium?

If you want to take the log base b of x, enter it as log(x, b).

To get log, you can either type it in or press diamond 7.

Nov 23, 2010 | Texas Instruments TI-89 Calculator

1 Answer

I am trying to figure out how to enter negative logs into the TI-30XA. I have a chemistry exam on Tuesday and we need to be able to do problems with the negative log and I can't seem to find any...

Use the +- key to the right of the decimal point to enter negative values and to change the sign of an existing value. For your sample:
2 EE +- 3 LOG +- =

Nov 15, 2010 | Texas Instruments TI-30XA Calculator

1 Answer

How do I calculate pH from H+ concentration on a ti 30a? [H+] 1.2 x 10 -5 M = ? pH = 4.9 I worked out pH to concentration but struggling for the opposite figured it out once and...

You calculate the pH with the relation
pH=-log(c) where c is the value of concentration in mol/L.
ex: c[H+]= 3.57x10^(-6)
pH: (-) [LOG] 3.57 [2nd][EE] [(-)] 6 [ ) ]
Here you have 2 change sign key press [(-)]

If you have the pH, and want to calculate the concentration that yields that value of the pH you proceed as follows
c= 10^(-pH)
Ex pH=8.23
c: [2nd][LOG] [(-) 8.23 [)]

General shortcut.
If concentration is of the form c[H+]= 1*10^(-a) where a is an integer between 0 and 14, the pH is equal to the positive value of exponent.

Ex:c[H+]=1x10^(-5) Mol/L then the pH is just the positive value of the exponent, ie 5 because pH=-log(1.x10^-5) =- log1 -log(10^-5)=0 -(-5)=5
Above I use the rule log(axb)=log(a)+log(b) , the rule log(a^b) = b*log(a)and the facts that log(1)=0 and log(10)=1, the last being true if log stands for log in base 10.

May 20, 2010 | Texas Instruments TI-30XA Calculator

1 Answer

How to use antilog?


As you well know the pH is the negative of the log in base 10 of the H+/H3O+ ion concentration. If we use [H+] to represent that concentration, then pH=-log[H+].
To obtain the [H+] you need to calculate the antilog. You write the definition in the form log[H+] =-pH and then calculate 10 to the power of each member. The equality remains valid as both members are treated similarly. Thus
10^( log[H+] ) = 10^(-pH)
Since raising 10 to a power is the inverse function of taking the log in base 10, 10^(log(x))=log(10^(x)) = x (they are inverse of one another), you are left with


Your calculator has a function [10 to x] accessed by pressing the [2nd] function key. To use it you must enter the negative value of the pH, press the [2nd] function key then the [10 to x], then the = key to get the result (concentration)
Exemple: let the pH=5.5, what is the H+ concentration?
With [(-)] being the change sign key, then
[H+]:[ (-) ] 5.5 [2nd][10 to x] [=]
The result is 0.000003162 or 3.16 x 10^(-6)

Hope it helps.

Dec 07, 2009 | Texas Instruments TI-30XA Calculator

2 Answers

I need help with a change of base formula problem.

It's simple here some logs:

log x base y = log x / log y
log (x*y) = log x + log y
log (x/y) = log x - log y
you just need to know that the default base of log in calculators is 10, so log x mean
log x base 10 = log x / log 10,
so write log x / log y for log x base y example:
log 8 base 2 = log 8 / log 2 that is how you write it on calculator, hope it was helpful

Jul 01, 2009 | Texas Instruments TI-83 Plus Calculator

3 Answers

How I can get the log base 10 number in a TI 89

"♦" "7" will give you log base 10 for the texas instuments TI-89 Titanium

Feb 20, 2009 | Texas Instruments TI-89 Calculator

1 Answer

Antilog in chemistry problem

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


Answer 0.001

Apr 22, 2008 | Texas Instruments TI-30XA Calculator

5 Answers

Ti89 titanium


Apr 02, 2008 | Texas Instruments TI-89 Calculator

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