Hi....

I need to calculate log2(1+100)

I know the result is 6.65, but I do not get it! help

I have finally worked out how to do log to the base 2 on my FX-83ES after nearly 2 years.

I hope yours is a similar problem.

The calculator must be in MATH mode or it will not work.

press the log button as the useless manual says then 2 then the right arrow on the replay button and then the number you want the log2 off..

Hey presto a correct answer

Posted on Apr 26, 2009

hi

to explain what a logarithm is, it's to involved to explain here but check sosmath.com for a detailed explaination of it

i use it to find root and powers of numbers. it turn up in certain natural process like radioactive decay, population growth, interest calculations..etc

for square rooting a number

square root of 16

type 16 press the log ( I'm using a scientific calculator a TI 30s to be exact) key to get 1.204119983 for a result divide by 2 to get .602059991 press shift or 2nd log key ( to get the 10^ key - this the anti- log key) to get 4 as a result

I hope this helps

to explain what a logarithm is, it's to involved to explain here but check sosmath.com for a detailed explaination of it

i use it to find root and powers of numbers. it turn up in certain natural process like radioactive decay, population growth, interest calculations..etc

for square rooting a number

square root of 16

type 16 press the log ( I'm using a scientific calculator a TI 30s to be exact) key to get 1.204119983 for a result divide by 2 to get .602059991 press shift or 2nd log key ( to get the 10^ key - this the anti- log key) to get 4 as a result

I hope this helps

May 25, 2014 | Texas Instruments TI 30XIIS Scientific...

Pressing 2 0 0 LN = will give you the natural logarithm of 200. If you need the common (base 10) logarithm, divide the result by the log of 10.

( 2 0 0 LN ) / ( 1 0 LN ) =

will give you the common log of 200.

( 2 0 0 LN ) / ( 1 0 LN ) =

will give you the common log of 200.

Jan 05, 2011 | Texas Instruments BA-II Plus Calculator

1 EE 14 LOG / 2 +<>- =

Jun 25, 2010 | Texas Instruments TI-30XA Calculator

You need the [EE] key to enter powers of 10 exponents.

You need the change sign key [+/-] key.

You need the [LOG] key.

pKa: 6.5 [EE] [+/-] 5 [LOG] [+/-]

There are two change sign: The first (either before or after the exponent 5), the last is entered after you press the log key , so as to negate (make negative if result is positive, and make positive is result is negative) the value.

The result should be 4.1870866 or 4.19

You need the change sign key [+/-] key.

You need the [LOG] key.

pKa: 6.5 [EE] [+/-] 5 [LOG] [+/-]

There are two change sign: The first (either before or after the exponent 5), the last is entered after you press the log key , so as to negate (make negative if result is positive, and make positive is result is negative) the value.

The result should be 4.1870866 or 4.19

Mar 08, 2010 | Texas Instruments TI-36 X Solar Calculator

I have no stake in the outcome of the calculation, but the way you wrote the expression is ambiguous (in my opinion). Look at the screen and see if that is what you meant. Result is displayed. Nevermind the calculator from which I grabed the screen, that is an irrelevant detail.

I notice that you have log(10), and wondered why you failed to take advantage of the fact that log(10)=1. Had you taken advantage of that fact, you could have (log(10))^6010 =1.

Now look athe next screen capture

If the last part of your expression is log(10^6010), a calculator much more sphisticated that yours was not able to handle the calculation, and found an infinite result but the result is finite. If you use the rule of the common logs log(10^6010) = 6010.

I am merely trying to convey the idea that any expression must be simplified as much as possible before typing it.

Now I make use of the rule about the logs

On the last screen, (log(10))^6010 is set equal to 1. The two expressions yiels identical results.

Once you use the rules about the logs, you expression becomes very easy to handle: You have one power to calculate. For that power you use the [^] key, below the [CLEAR] key.

I think that what I showed you above should help you figure out how to enter the expression on your calculator.

I notice that you have log(10), and wondered why you failed to take advantage of the fact that log(10)=1. Had you taken advantage of that fact, you could have (log(10))^6010 =1.

Now look athe next screen capture

If the last part of your expression is log(10^6010), a calculator much more sphisticated that yours was not able to handle the calculation, and found an infinite result but the result is finite. If you use the rule of the common logs log(10^6010) = 6010.

I am merely trying to convey the idea that any expression must be simplified as much as possible before typing it.

Now I make use of the rule about the logs

On the last screen, (log(10))^6010 is set equal to 1. The two expressions yiels identical results.

Once you use the rules about the logs, you expression becomes very easy to handle: You have one power to calculate. For that power you use the [^] key, below the [CLEAR] key.

I think that what I showed you above should help you figure out how to enter the expression on your calculator.

Feb 10, 2010 | Texas Instruments TI-84 Plus Calculator

Hi,

Ex: calculate -log(15.32)

Here is a picture of the change sign

Hope it helps

Ex: calculate -log(15.32)

- Calculate the log 15.32 [LOG] [=] result 1.185258765
- Then press the change sign key (-) result -1.185258765

Here is a picture of the change sign

Hope it helps

Dec 13, 2009 | Texas Instruments TI-30XA 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,

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

Hello

When you talk about the natural logarithms [LN], the inverse of the log is the exponential

ln (exp(x))=exp(ln(x)) =x

but with othe logarithms, thinhs are different

Let us take

**y=10^(x)**

then with log standing as usual for log in base 10 take the log of both members in the above equality

log(y) =log{10^(x)) = x* log(10) : this is a property of the log function.

However, log(10) =1 This is also a property of the log

Gathering results, we end up with the new equality

log(y)=x or x=log(y)

Without entering into extraneous details that mathematicians adore, we can state that there is an equivalenve between y=10^(x) and x=log(y)

**y=10^(x) <-->x=log(y)** where <--> means implication in both directions.

Here is a screen capture from another calculator to convinve you. Your calculator cannot do that.

Hope it helps.

When you talk about the natural logarithms [LN], the inverse of the log is the exponential

ln (exp(x))=exp(ln(x)) =x

but with othe logarithms, thinhs are different

Let us take

then with log standing as usual for log in base 10 take the log of both members in the above equality

log(y) =log{10^(x)) = x* log(10) : this is a property of the log function.

However, log(10) =1 This is also a property of the log

Gathering results, we end up with the new equality

log(y)=x or x=log(y)

Without entering into extraneous details that mathematicians adore, we can state that there is an equivalenve between y=10^(x) and x=log(y)

Here is a screen capture from another calculator to convinve you. Your calculator cannot do that.

Hope it helps.

Oct 06, 2009 | Texas Instruments TI-30 XIIS Calculator

Feb 08, 2014 | Casio FX-250HC Calculator

Aug 06, 2013 | Casio FX-250HC Calculator

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