I DONT KNOW HOW TO ENTER LOGS INTO A CALCULATOR WHEN IT DOESNT HAVE A BASE OF 10

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Posted on Jan 02, 2017

The 30XA only has logarithm functions for base e and base 10. However, you can calculate the logarithm to any base by using the relationship logb x = log x / log b = ln x / ln b

To calculate log3 of 9, press

9 LOG / 3 LOG =

or

9 LN / 3 LN =

To calculate log3 of 9, press

9 LOG / 3 LOG =

or

9 LN / 3 LN =

Apr 03, 2011 | Texas Instruments TI-30XA Calculator

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.

Nov 15, 2010 | Texas Instruments TI-83 Plus Calculator

Press LOG for common (base-10) logarithm or LN

Jul 04, 2010 | Texas Instruments TI-83 Plus Calculator

The TI-83 doesn't have such a function built-in. Use the relationship logb(x) = log(x)/log(b) = ln(x)/ln(b).

May 13, 2010 | Texas Instruments TI-83 Plus 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,

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

Hello,

If you know the theory skip this and go to**Application**

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**

**Application**: What is the antilog of 3.76?

antilog of 3.76 =10^(3.76) = 5754.399373

Take the log in base 10 of this number and you recover 3.76

You enter it as follows

10[^]3.76[ENTER/=] gives 5754.399373

And log(5754.399373)= 3.76

Hope it helps

If you know the theory skip this and go to

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

If

antilog of 3.76 =10^(3.76) = 5754.399373

Take the log in base 10 of this number and you recover 3.76

You enter it as follows

10[^]3.76[ENTER/=] gives 5754.399373

And log(5754.399373)= 3.76

Hope it helps

Jun 13, 2009 | Sharp EL-531VB Calculator

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

Feb 20, 2009 | Texas Instruments TI-89 Calculator

To get an exponent press 2nd then LN or LOG

LN is natural logs, so entering 1 then 2nd LN gives you the natural exponent of 1 which is.2.71828....pressing LN gives you the natural log of 2.71828.... which is 1

LOG is logs to base 10, so entering 1 then 2nd LN gives you the 10 to the power 1 which is.10, pressing LOG gives you the log of 10 which is 1 again

LN is natural logs, so entering 1 then 2nd LN gives you the natural exponent of 1 which is.2.71828....pressing LN gives you the natural log of 2.71828.... which is 1

LOG is logs to base 10, so entering 1 then 2nd LN gives you the 10 to the power 1 which is.10, pressing LOG gives you the log of 10 which is 1 again

Dec 04, 2008 | Texas Instruments TI-30XA Calculator

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