Question about Casio FX-260 Calculator

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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

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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

If you want to take the log of X using base B, where B is an arbitrary number, use the following equation:

logB(X) = log10(X) / log10(B)

In words this is the log base B of X equals the log base 10 of X divided by the log base 10 of B. This equation can now be solved using your calculator.

I hope this helps you out.

logB(X) = log10(X) / log10(B)

In words this is the log base B of X equals the log base 10 of X divided by the log base 10 of B. This equation can now be solved using your calculator.

I hope this helps you out.

Feb 09, 2011 | Vacuums

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

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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