Question about Texas Instruments TI-83 Plus Calculator

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Posted on Jan 02, 2017

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hi

if it a scientific calculator there should be a surd key ( look like a square root symbol, other way to get a square look for the "^" key that the exponent or power key do the follow type in the number be square rooted and press the "^" key and type "(1/2)" ( without the "" of course) and you will get the square root of your number. You can use the log and ln keys as well

type in your number you want square rooted and press the log or ln keys divide the result by 2 press the anti log key use might have press the 2nd key or shift key and press the log ( or 10^ key) or ln( or e^ key) the 10^ key and the e^ key change log to normal numbers

i.e.

square root 9

type 9 press log to get .954242509 for a result

divide by 2 to get .477121255 press the 10^ key to get

3 for a answer.

if it a scientific calculator there should be a surd key ( look like a square root symbol, other way to get a square look for the "^" key that the exponent or power key do the follow type in the number be square rooted and press the "^" key and type "(1/2)" ( without the "" of course) and you will get the square root of your number. You can use the log and ln keys as well

type in your number you want square rooted and press the log or ln keys divide the result by 2 press the anti log key use might have press the 2nd key or shift key and press the log ( or 10^ key) or ln( or e^ key) the 10^ key and the e^ key change log to normal numbers

i.e.

square root 9

type 9 press log to get .954242509 for a result

divide by 2 to get .477121255 press the 10^ key to get

3 for a answer.

Aug 12, 2014 | Texas Instruments TI-15 Calculator

The inverse of the log function is the function 10^

y=log(x) has the function x=10^y as inverse.

y=ln(x) has the function x=e^y

log and 10^ have the same key.

ln and e^ have the same key.

y=log(x) has the function x=10^y as inverse.

y=ln(x) has the function x=e^y

log and 10^ have the same key.

ln and e^ have the same key.

Jul 15, 2014 | Office Equipment & Supplies

Press LOG 3 ENTER.

The LOG key on the keyboard gives you base-10 (common) logs while the LN key gives you base-e (natural) logs. If you want the log to another base you can use the logBASE function found in the catalog.

The LOG key on the keyboard gives you base-10 (common) logs while the LN key gives you base-e (natural) logs. If you want the log to another base you can use the logBASE function found in the catalog.

Apr 07, 2014 | Texas Instruments TI-84 Plus Calculator

What you call the antilog is also known as as the power function. More specifically, the inverse of a decimal log is the function 10 to a power, (10^x) It shares the same physical key as the log function. If the calculator has a key marked [LOG] you have to press the [2nd or SHIFT] key followed by the [LOG] key to activate the anti-log.

For natural log (logarithms in base e) the antilog is the exponential function (e^x). It shares the same physical key as the [LN] function.

For natural log (logarithms in base e) the antilog is the exponential function (e^x). It shares the same physical key as the [LN] function.

May 09, 2011 | Texas Instruments TI-30XA Calculator

(-) LOG 1 . 7 5 EE (-) 5 ENTER

The (-) is just to the right of the decimal point, the LOG key is in the leftmost column between the GRAPH and LN keys, the EE key is just above the 7 key.

The (-) is just to the right of the decimal point, the LOG key is in the leftmost column between the GRAPH and LN keys, the EE key is just above the 7 key.

Dec 11, 2010 | Texas Instruments TI-85 Calculator

Hello,

This post answers two questions

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

**[H+]=10^(-pH)**

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)

**Example**s

1. 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)

Calculating the pH

Shortcut:

For all H+/H3O+ concentrations of the form**1.*10^(a)** where a is** an integer number between 0 and -14**, the pH is the negative value of the exponent.

Concentration =10^(-3), pH=3

Concentration=10^(-11), pH=11

For other concentrations such as 3.567*10^(-8), one cannot use the shortcut above, but have to calculate the log of the concentration

[H+/H3O+] = 3.567*10^(-8)

pH= - log(3.567*10^(-8))

This is keyed in as follows (to minimize the number of parentheses)

**8 (-) [2nd][10 to x] [*] 3.567 [LOG] [=] (-)**

Here you have two (-) change sign, the first is entered after the exponent of 10, the other at the end of the calculation to take the negative of the displayed result.**You may notice that it is entered in the reverse order of the defining relation **- log(3.567*10^(-8)).

To verify your calculation, the result is 7.447696891 or just 7.45

If you have a problem with the first (-) try entering it before you type in 8.

Hope it helps** **and thank you for using FixYa

And please, show your appreciation by rating the solution**.**

This post answers two questions

- How to obtain the concentration knowing the pH?
- How to obtain the pH knowing the concentration

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

To obtain the [H+] you need to calculate the antilog. You write the definition in the form

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

1. Let the pH=5.5, what is the H+ concentration?

With [(-)] being the change sign key, then

[H+]:

Calculating the pH

For all H+/H3O+ concentrations of the form

Concentration =10^(-3), pH=3

Concentration=10^(-11), pH=11

For other concentrations such as 3.567*10^(-8), one cannot use the shortcut above, but have to calculate the log of the concentration

[H+/H3O+] = 3.567*10^(-8)

pH= - log(3.567*10^(-8))

This is keyed in as follows (to minimize the number of parentheses)

Here you have two (-) change sign, the first is entered after the exponent of 10, the other at the end of the calculation to take the negative of the displayed result.

To verify your calculation, the result is 7.447696891 or just 7.45

If you have a problem with the first (-) try entering it before you type in 8.

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

Hello,

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

**[H+]=10^(-pH)**

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.

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

To obtain the [H+] you need to calculate the antilog. You write the definition in the form

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

Exemple: let the pH=5.5, what is the H+ concentration?

With [(-)] being the change sign key, then

[H+]:

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

Hello,

**Please read the following with patience.**

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

Let's get back to you problem : To obtain the antilog you use the key marked with a 10^x.

Exemple1: What is the antilog of 3

Answer: Antilog(3) =10^3 =1000 because log(1000) =3.

Exemple 2: What is the antilog of 1.543765?

Antilog(1.543765) = 10^ (1.543765)= 34.97558603

Exemple3: What is the antilog of -0.37654?

Answer: Antilog (-0.37654) =10^(-0.37654) =0.4202038237

Look on the body of the calculator if there is key or a [Shift] key labeled with 10^x ( a 10 with a small raised x). If you are able to find it, use it. If not, you can always use the [raise to the power key] ( a caret ^, or Y to the x or X to the y.) For the last exemple, you type it as follows.

10 [^] (-) 0.37654 [=/ENTER]. The (-) is the change sign not the regular Minus.

Hope it helps.

Thanks for using FixYa.

- Let y=10^(x) ( 10 raised to the power of x)
- Take the log of both tems of the equality.
- You getlog(y)=log[10^(x)] where I used square brackets for clarity.
- But from the general properties of logarithmslog(b^(a)) = a*log(b)
- Applied to our expression above ( in Number 1)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

Let's get back to you problem : To obtain the antilog you use the key marked with a 10^x.

Exemple1: What is the antilog of 3

Answer: Antilog(3) =10^3 =1000 because log(1000) =3.

Exemple 2: What is the antilog of 1.543765?

Antilog(1.543765) = 10^ (1.543765)= 34.97558603

Exemple3: What is the antilog of -0.37654?

Answer: Antilog (-0.37654) =10^(-0.37654) =0.4202038237

Look on the body of the calculator if there is key or a [Shift] key labeled with 10^x ( a 10 with a small raised x). If you are able to find it, use it. If not, you can always use the [raise to the power key] ( a caret ^, or Y to the x or X to the y.) For the last exemple, you type it as follows.

10 [^] (-) 0.37654 [=/ENTER]. The (-) is the change sign not the regular Minus.

Hope it helps.

Thanks for using FixYa.

Nov 19, 2009 | Sharp EL-531VB Calculator

Hello,

Why complicate matters for yourself?

You were asked to calculate log(6.02x10^23), let your finger do the calculating.

I checked the claculator manual and you have to enter it all at one go. Pay attention to the key strokes. You will not do it in such a laborious manner

23[10^x] [=] gives 10^23

23[10^x]*6.02 [=] multiplies 6.02 by 10^23 to give you Avogadro s number.

To find its logarithm in base 10, you have to enter the number then press [LOG] . When you press [LOG] you are calculating the log of the last result. You obtain 23.77959649.

If I had given you the key strokes directly you might have not understood why I do things this way. Now the actual key strokes you enter

**23[10^x] * 6.02 [LOG] [=]**

If you are only looking for your result, you are done. You can ignore what follows.

If you know the rules for the logarithms, you can do the calculation more easily.**log** here is **log in base 10**

Rule 1** Log(a*b)= log(a) + log(b)**

Rule 2** Log(c^n) = n* log(c)**

Rule 3** log(10)=1**

Thus applying the rules

log(6.02*10^23) = log(6.02) + log(10^23) first rule applied

= log(6.02) + 23*log(10) 2nd rule applied

= log(6.02) + 23 3rd rule applied

= 0.77959649 + 23= 23.77959649

Hope it helps

Why complicate matters for yourself?

You were asked to calculate log(6.02x10^23), let your finger do the calculating.

I checked the claculator manual and you have to enter it all at one go. Pay attention to the key strokes. You will not do it in such a laborious manner

23[10^x] [=] gives 10^23

23[10^x]*6.02 [=] multiplies 6.02 by 10^23 to give you Avogadro s number.

To find its logarithm in base 10, you have to enter the number then press [LOG] . When you press [LOG] you are calculating the log of the last result. You obtain 23.77959649.

If I had given you the key strokes directly you might have not understood why I do things this way. Now the actual key strokes you enter

If you are only looking for your result, you are done. You can ignore what follows.

If you know the rules for the logarithms, you can do the calculation more easily.

Rule 1

Thus applying the rules

log(6.02*10^23) = log(6.02) + log(10^23) first rule applied

= log(6.02) + 23*log(10) 2nd rule applied

= log(6.02) + 23 3rd rule applied

= 0.77959649 + 23= 23.77959649

Hope it helps

Oct 10, 2009 | Texas Instruments TI-30XA Calculator

Hi ;

for log base 10 use the the "log" key

example :

log(2) + log(3) = log(6)

enter 2 press the log key you will get 0.30102999 press the '+" key than 3 log you should get 0.47712125 for it Press the "=" or enter and you should get 0.778151250 for a answer than press inv or 2nd log and you should get 6 for a answer

and that all there is to it

for log base 10 use the the "log" key

example :

log(2) + log(3) = log(6)

enter 2 press the log key you will get 0.30102999 press the '+" key than 3 log you should get 0.47712125 for it Press the "=" or enter and you should get 0.778151250 for a answer than press inv or 2nd log and you should get 6 for a answer

and that all there is to it

Oct 07, 2009 | Texas Instruments BA-II Plus Calculator

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