How do I find the log button or how do I use the calculator to calculate a log problem? i.e. (log (learning curve))/(log(2))

Press Y=
On the Y1 line enter the expression LOG(X+2)+LOG(X+3) by pressing
LOG ALPHA [X] ) + LOG ALPHA [X] + 3 )
On the Y2 line enter the expression 7-X by pressing
7 - ALPHA [X]
Press GRAPH to see the graph. To see where they intersect, press 2ND [CALC] 5. See "First curve?" in the lower left. Use up/down-arrow to select one of the two curves and press ENTER. See "Second curve?" in the lower left. Use up/down-arrow to select the other curve and press ENTER. See "Guess?" in the lower left. Press ENTER. The calculator will show the coordinates of the intersection.

You cannot be serious. The digit keys allow you to enter the digits that make up the number you are using in your calculations. +, x, - / allow you to perform an addition, a multiplication, a subtraction, or a division. Keys with x^2, sin, tan, cos, log, allow you to calculate the relevant functions.
You get the idea.
To learn how to use the calculator, read the manual. At least skim through it looking for instructions on how to use a particular task. There is no shortcut.

For the inverse natural log, press 2nd LN. For the inverse common log, press 2nd LOG.

For example, to calculate the inverse natural log of 2, press 2nd LN 2 ENTER and you'll get about 7.389 .

Graph the function then use the trace function to locate the point(s) where two curves intersect. In your case one of the curves is the X-AXIS.

See the screen capture to learn how to use the trace functionality.

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
log4[(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.

Hello,
To raise a number to some power other than 2 or 3, you use the [y to the x] key. Exemple 2[y to the x] 5 = 2x2x2x2x2=32.
Hope it helps.

PS. I could not understand what you wanted with those expressions.

Sorry, this might not be much help but apparently the calculator is set in a mode called "base 10".  I have never learned logrithms really in school, but did a little for my ACT.  I figured that in order to do a problem, you need to divide.  EXAMPLE: 8 log 2 = ? would be entered [log] [8] / [log] [2] [EXE] and the answer 3 will be displayed.  Not much help, but maybe this will get you in the right direction. 

Look the calculators has a default base 10 and you can't change it but you may learn this:
log x base y = log x / log y
log (x*y) = log x + log y
log (x/y) = log x - log y
so instead input log 8 base 2 you need to input log 8 / log 2
your example answer is about 0.919... if i get right what did you write.
Glad to help :)

Hello,
From Wikipedia.
Mathematically the experience curve is described by a power law function sometimes referred to as Henderson's Law:

• [4]

where

• is the cost of the first unit of production
• is the cost of the nth unit of production
• is the cumulative volume of production
• is the elasticity of cost with regard to output[

If that is the law involved you enter it as follows

60000*500[ ^ ] [(-)] 0.415 [=] result should be 4550.705

Here
[ ^ ] is the power key, between x-square and log
[(-)] is the change sign (or negation) key in the leftmost column, just above [RCL]

Hope it helps.

Hello, The first remark is this. The ratio log(0.8)/log(2) is by definition binary log of 0.8 ( meanibg log in base 2 of 0.8) The second remark is good news for you.: When in MathIO, [SHIFT][MODE][1:MathI0], your calculator can compute directly the log in any base of any number. To use the functionality, you press the key below [ON]. It is a log symbols with a small solid square, and and erect empty rectangle.
When you press the key the screen shos a template. In the smaller and lowered square goes the base (here 2). When you enter the base, press the right arrow to move cursor to the rectangle inside the parentheses and type in 0.8. Use the right arrow to move the cursor out of the parentheses.
Practice: try that. the result should be -0.321928095. Good. Now use the replay button to move cursor to the command line outside the right parenthesis. I am going to give you the other key stokes. Let log(2,0.8) be my representation for log in base2 of 0.8. Then you enter the rest of the calculation as
log(2, 0.8) [*] [ ( ] [(-)] 0.321 [ ) ] [*] 2500 [=]
Here [*] is the multiplication sign [(-)] is the negation (change sign), just above the [RCL] key.
[ ( ] is the left parenthesis and [ ) ] its matching right parenthesis. Both of them enclose the number -0.321. It underlined them for clarity.
The result should be 259.0716344
If you do not like what I did above, you can always do things the old way. You use the log key just above the [COS] key. In that case -0.3219*[LOG] (0.80)/[LOG] 2 * 2500. Do not to supply the ) for both log. The calculator inserts the left ones by default.
Hope it helps.