I don't get it but here some help with logs:

log x base y = log x / log y (when log x is default base 10 in calculators) and

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

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

the bases is default 10

hope it's helped

Posted on Jul 06, 2009

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Posted on Jan 02, 2017

log (k_a)=-4.7

-log(k_a)=4.7

-log(k_a)=0 - log(k_a)=log(1)-log (k_a)=

=**log(1/k_a**)**=4.7**

**Raise both members of the last equality to the power of 10**

10^(log(1/k_a))=**1/k_a**=10^(4.7)

**k_a=10^(-4.7)=1.99526*10^(-5)**

-log(k_a)=4.7

-log(k_a)=0 - log(k_a)=log(1)-log (k_a)=

=

10^(log(1/k_a))=

Mar 22, 2014 | Office Equipment & Supplies

Is the original problem (log 10,000 - log 1,000) / (4 log 1.0125) ? If so, press

( 1 0 0 0 0 LOG - 1 0 0 0 LOG ) / ( 4 * 1 . 0 1 2 5 LOG =

and you should see the expected result.

( 1 0 0 0 0 LOG - 1 0 0 0 LOG ) / ( 4 * 1 . 0 1 2 5 LOG =

and you should see the expected result.

Jul 25, 2011 | Office Equipment & Supplies

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

Sorry the answer comes too late for your final exam. If you took the time to really understand what the inverse of a log function is, you would have saved yourself the anxiety before the test.

Anyway, the inverse of the logarithm in base 10 IS the power function with base 10.

y=log(x) <=> x=10^(y).

Some calculator manufacturers use the marking ^-1 to represent an inverse function. But recently one sees new calculators with the new notation.

By the way, the inverse of the natural log function (LN) is the exponential function (e^(x))

Anyway, the inverse of the logarithm in base 10 IS the power function with base 10.

y=log(x) <=> x=10^(y).

Some calculator manufacturers use the marking ^-1 to represent an inverse function. But recently one sees new calculators with the new notation.

By the way, the inverse of the natural log function (LN) is the exponential function (e^(x))

May 02, 2011 | Casio fx-300ES Calculator

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

Assuming you want the common antilog of 11/20:

2nd 10^x 1 1 / 2 0 ENTER

10^x is on the LOG key, to the left of the 7 key.

2nd 10^x 1 1 / 2 0 ENTER

10^x is on the LOG key, to the left of the 7 key.

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

1 EE 14 LOG / 2 +<>- =

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

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

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

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