The natural log button, ln, is in the third row from the top.
go to 2nd LN and if you just want "e" go to the first power
Assuming you mean 'logarithm to the base 'e' ( natural logarithm. On the calculator its symbol is 'ln'. Hence ;ln 2 = 0.69314718....
To find ln 2.33, you need a calculator. It is the solution of the equation e^x = 2.33. ln 2.33 = 0.84586 (using a calculator)
It depends. If you mean (ln e)7, then the answer is 1, since (ln e) = 1. If you mean ln(e7), then the answer is 7, since ln(e7) = 7 (ln e) and (ln e) = 1.
a calculator adds, subtracts, multiplies and divides numbers and numbers with decimals. it stores single numbers to memory, solves percentages, and square roots. a scientific calculator does a little more than a calculator. it includes the Greek letter or the numeral pi, finds exponents, roots, plots data, contains the trigonometric symbols sin, cos, and tan, and the logarithmic symbols ln, log, X^10 and X^e
it's an expression, so you'll have to use a calculator.=========================Right! In this day and age, we've finally reached the point where nobodyknows how to do squat without his machines. Plus ... nobody has a cluewhether the machine has given him the right answer. Beautiful!ln(5) - 3 ln(2) =ln(5) - ln(23) =ln (5/23) = ln (5/8).Note:That's not the 'solution' to anything. It's just a different, neater way of writingthe original expression that's in the question. In an algebra class, the answerdoesn't matter. All they really want is for you to learn how to get this far, andno calculator will tell you that it's ln(5/8).If your class is so regimented that you need the numerical answer, then you cango ahead and use the calculator for that:ln (5/8 ) = -0.47 (rounded)
(ex-e-x)/2=8 ex-e-x=16 ex = 16-e-x ln(ex) = ln(16-e-x) x = ln(16-e-x) x = ln((16ex-1)/ex) x = ln(16ex-1) - ln(ex) x = ln(16ex-1) - x 2x = ln(16ex-1) e2x = eln(16e^x - 1) e2x = 16ex-1 e2x-16ex +1 = 0 Consider ex as a whole to be a dummy variable "u". (ex=u) The above can be rewritten as: u2-16u+1=0 u2-16u=-1 Using completing the square, we can solve this by adding 64 to both sides of the equation (the square of one half of the single-variable coefficient -16): u2-16u+64=63 From this, we get: (u-8)2=63 u-8=(+/-)sqrt(63) u=sqrt(63)+8, u= 8-sqrt(63) Since earlier we used the substitution u=ex, we must now use the u-values to solve for x. sqrt(63)+8 = ex ln(sqrt(63)+8)= ln(ex) x = ln(sqrt(63)+8) ~ 2.769 8-sqrt(63) = ex ln(8-sqrt(63)) = ln(ex) x = ln(8-sqrt(63)) ~ -2.769 So, in the end, x~2.769 and x~-2.769. When backsubbed back into the original problem, this doesn't exactly solve the equation. Using a graphing calculator, the solution to this equation can be found to be approximately x=2.776 by graphing y=ex-e-x and y=16 and using the calculator to find the intersection of the two curves. This is pretty dang close to our calculated value, and rounding issues might account for this difference. The calculator, however, suggests that x~-2.769 is not a valid solution. This makes sense, and in fact it isn't a valid solution if you look at the graphs. This is an extraneous answer.
To enter a natural log, press the LN button. To enter a log with base 10, press the LOG button. To enter a log with a base other than those, divide the log of the number with the log of the base, so log6(8) would be log(8)/log(6) or ln(8)/ln(6). (The ln is preferred because in calculus it is easier to work with.)
in math, ln means natural log, or loge and e means 2.718281828
I think you are thinking "natural logarithm" which is ln (lowercase L, not I). If you have taken calculus you learn about logarithm and its relationship with exponents
Do you mean ln(x-2), or ln(x)-2? If it is ln(x-2): 1/(x-2) If it is ln(x)-2: 1/x