Question about Casio FX-260 Calculator

Here something that would help you:

log x base y = log x / log y

log (x*y) = log x + log y

log (x/y) = log x - log y

so log 24 base 2 it's log 24 / log 2 etc.

and if i get right what you wrote x= 10 ^ ( 3 * (log 5) ) if no inverse it it's easy really Hope it's helped you.

Posted on Jul 06, 2009

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Posted on Jan 02, 2017

Type **solve(log(x^5,16)-log(125,64)+log(x^(1/2),4),x,0.5**

Its equation with guess 0,5. Then press enter and you will give result

x=5^(1/3)=1.70997

Or, you can use equation solver with eqn=0(MATH, then B option)

See captured images below

x=5^(1/3)=1.70997

Or, you can use equation solver with eqn=0(MATH, then B option)

See captured images below

Feb 26, 2013 | Texas Instruments TI-83 Plus Calculator

In equation mode, you have system of linear equations (3 unknown) you have polynomial (quadratic and cubic), and solver. Use the solver foe any type of equation (nonlinear, polynomial of order higher than 4, trigonometric, exponential, logs).

May 21, 2012 | Casio Scientific Calculator Fx-570 Fx570...

Use the "log" key. For example, to calculate the log base 10 of 2, press

"log" "2" =

"log" "2" =

Mar 29, 2012 | Casio FX991ES Scientific Calculator

Of course you can divide by a log. As an example, to divide 2 by the log of 3, press

2 / 3 LOG =

To divide 2 by the natural log of 3, press

2 / 3 LN =

2 / 3 LOG =

To divide 2 by the natural log of 3, press

2 / 3 LN =

Oct 27, 2011 | Texas Instruments TI-30XA Calculator

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.

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 17, 2011 | Texas Instruments TI-30XA Calculator

The 30XIIS does not have the capability of solving equations. It can, however, calculate expressions using logarithms. Use the LOG key for common logarithms and the LN key for natural logarithms. In both cases, press the function key, enter the value or expression, and then ) to match the ( the function automatically enters.

The LOG and LN keys are the second and third keys from the top in the leftmost column of the keyboard.

The LOG and LN keys are the second and third keys from the top in the leftmost column of the keyboard.

Nov 13, 2010 | Office Equipment & Supplies

Let us start with the definitions:

Let c be the concentration of H+/H30+ ions in the the solution (expressed in mol/L).

By definition the pH =-log(c)

Rewriting the equation as log(c)=-pH

Taking both members of the last equation to the power of 10

10^(log(c))=10^(-pH)

Since 10^(log(x))=log(10^x)= x (identity for inverse functions)

we get c=10^(-pH).

To handle concentartion-pH problems, you have two relations, namely

pH=-log(c) : to calculate pH from the concentrartion

and

c=10^(-pH): to calculate the concentration from the pH

A SHORTCUT WORTH REMEMBERING.

If you concentration is of the form c=1.*10^(-a) where a is between 0 and 14, the pH is simply the value of a.

Ex: c=1.*10^(-3), pH=3

c=1*10^(-8), pH =8.

Now let us take a more general example: c=6.54*10^(-9), pH=?

pH=-log(6.54*10^(-9))

You have to enter twice the change-sign [(-)], the one on the bottom row of the key pad.

Here is a screen capture from a TI84Plus calculator.

As you can see, you can use the general power key [^] to enter 10^(-9)

or the shortcut [2nd][ ' ] to access the (EE) which appears on screen as E. The second way is more efficient (left screen)

On the screen to the right you will notice that I did not close the parentheses, yet I obtained the correct result. However, I suggest you always close parentheses to avoid syntax errors.

Let c be the concentration of H+/H30+ ions in the the solution (expressed in mol/L).

By definition the pH =-log(c)

Rewriting the equation as log(c)=-pH

Taking both members of the last equation to the power of 10

10^(log(c))=10^(-pH)

Since 10^(log(x))=log(10^x)= x (identity for inverse functions)

we get c=10^(-pH).

To handle concentartion-pH problems, you have two relations, namely

pH=-log(c) : to calculate pH from the concentrartion

and

c=10^(-pH): to calculate the concentration from the pH

A SHORTCUT WORTH REMEMBERING.

If you concentration is of the form c=1.*10^(-a) where a is between 0 and 14, the pH is simply the value of a.

Ex: c=1.*10^(-3), pH=3

c=1*10^(-8), pH =8.

Now let us take a more general example: c=6.54*10^(-9), pH=?

pH=-log(6.54*10^(-9))

You have to enter twice the change-sign [(-)], the one on the bottom row of the key pad.

Here is a screen capture from a TI84Plus calculator.

As you can see, you can use the general power key [^] to enter 10^(-9)

or the shortcut [2nd][ ' ] to access the (EE) which appears on screen as E. The second way is more efficient (left screen)

On the screen to the right you will notice that I did not close the parentheses, yet I obtained the correct result. However, I suggest you always close parentheses to avoid syntax errors.

Feb 11, 2010 | Texas Instruments Office Equipment &...

This is not a calculator problem. Use the rules you have learned to simplify your problem.

For a a log in any base.** log(a*b) =log(a) +log(b)** or log(a)+log(b) =log(a*b).

Thus

log(x+1)+log(x-2)= log[(x+1)*(x-2)]=1

Use the fact that log4=log in base 4 and rewrite the equation with the appropriate symbols.** log4[(x+1)*(x-2)] =1 **

However**1=log4(4)** and

l**og4[(x+1)*(x-2)]=log4(4)**

The equality of the two logs implies the equality of their arguments (contents) and

**(x+1)*(x-2) =4**

Now you solve this quadratic equation by the methods you must have learned (sorry I have to leave some thing for you to do, to ensure that proper learning is achieved).

Once you find the roots of the quadratic equation, verify that each term in your original expression has meaning-- x+1 positive and x-2 positive.

If one of the roots makes the argument (x+1) or (x-2) negative, reject it. because the argument of a log function cannot be negative.

For a a log in any base.

Thus

log(x+1)+log(x-2)= log[(x+1)*(x-2)]=1

Use the fact that log4=log in base 4 and rewrite the equation with the appropriate symbols.

However

l

The equality of the two logs implies the equality of their arguments (contents) and

Now you solve this quadratic equation by the methods you must have learned (sorry I have to leave some thing for you to do, to ensure that proper learning is achieved).

Once you find the roots of the quadratic equation, verify that each term in your original expression has meaning-- x+1 positive and x-2 positive.

If one of the roots makes the argument (x+1) or (x-2) negative, reject it. because the argument of a log function cannot be negative.

Dec 14, 2009 | Casio FX-300MS Calculator

In any scientific calculator log2(n) can be calculated with either ln or log function as
follows

Log2(n)= ln(n) / ln(2)

Or

Log2(n)=log(n) / log(2)

both will give nearly the same answers

Log2(n)= ln(n) / ln(2)

Or

Log2(n)=log(n) / log(2)

both will give nearly the same answers

Dec 08, 2007 | Casio FX-300MS Calculator

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