How do i perform the calculation of log (base 2) 8? The answer should be 3....that is it is the inverse calculation of 2 cubed =8. Can any one help with the log calculation?

By 'log inverse', you presumably mean the inverse of the logarithm function. There are two logarithm functions on most scientific calculators. Firstly [ln], or natural logarithms, to the base e, where the inverse is e^x. Secondly [log], or logarithms to the base 10, where the inverse is 10^x.

Example 1 : ln(2) = 0.69314718 so e^0.69314718 = 2
Example 2: log(2) = 0.301029995 so 10^0.30102995 = 2

The inverse of the log function is the power function.
For log in base 10 that inverse is 10 to a power of
More generally, let b be the base of the logarithm. If y=log_b (x) then x=b^y
For your case log=log_10, to calculate the inverse you perform 10^(-2)=0.01=1/100
On calculators the log in base 10 and its inverse share the same physical key. One is accessed directly, the other is the shifted key function.

Depending on the actual log function you are talking about, the inverse function (what you call the anti log) can be one of several things.
The inverse of the natural log (LN) is the exponential function e^x.
If y= ln(x) then x=e^y
On a calculator functions that are inverse of one another share the same physical key. If one function is marked on a key, the inverse function is usually the SHIFTED (or 2nd) key
The inverse of the common (decimal logarithm) is the power function with 10 as base.
Example: if y= log(x) then x=10^y
For an arbitrary base (must be positive) if y= log_b(x) then x=b^y.
Note: I do not think that the FX-300ms can perform log calculations in bases other than e (natural log) or 10 (common decimal log).

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))

fllow the following steps
To do inverse log (to the base 10), e.g. antilog 2 :
SHIFT ... 10^x ... 2 ... = (answer is 100)

To do inverse ln (to the base e), e.g. antiln 2 :
SHIFT ... e^x ... 2 ... = (answer is 7.389056...)

If you want to look at the whole manual, go here :
http://ftp.casio.co.jp/pub/world_manual/...
good luck, do rate the solution if got help

Hi,
Sorry to contradict you but there are many types of logarithms, the most important ones are

1. the common logarithms (log in base 10),
2. the natural logarithms (logarithms in base e)
3. the binary logarithms (logarithms in base 2)

On calculators, the common logarithms are labeled [LOG], and the natural logarithms are labeled [LN].

The inverse function of the natural log function is the exponential (e^(x)), and the inverse of the log in base ten function is the function ten to the power of. It is called (sometimes) the antilog

Ex:
Question What is the antilog of 3.5678?
Answer The antilog of 3.5678 is 10^(3.5678) = 3696.579068
Verification: log(3696.57908) =3.5678

Hope it helps.

Raise the base of your log to the power produced by the answer.

Example: If ln(x) = y, raise e (base of the natural logarithm) to the y to get x = e^y
Example 2: log(x) = y - x = 10^y

So if log(x) = 2, x = 10^2 = 100

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.

The problem is simple. You're trying to get an [H+] concentration which is obviously going to have a value of some number times ten raised to a negative power. Therefore, you have to insert the negative value of the pH into the 10^(x). When you insert said negative number you will come out with the right answer.
i.e.
The pH of a sample of human blood was calculated to be 7.41. What is the [H+] concentration of the blood?
10^(-7.41) = [H+]
[H+] = 3.9 E-8  
(the answer should only have two sig. figs because the pH has two digits after the decimal.

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