When ^ mean to the X power how do I calculate the -3 into the 1/n?
Need to know: a^(-b) = 1/(a^b) a^(x)/a^(y) = a^(x-y) If you have not mistaken any other parenthesis, (2n^-2) = 2/ (n^2) (1/n) ^ -3 = (1^-3) / (n^-3) = 1 / (n^-3) = n^3 therefore, after multiplying, the answer is 2 (n^3) / (n^2) = 2 (n^(3-2)) = 2 (n^1) = 2n Hope I helped!
You could use log tables, in which case log(1.1)=0.04139So 10*log(1.1)=0.4139So antilog(0.4139)=2.59358I got 2.5937 using calculator, while I got 2.59358 without calculator.Pen-&-Paper Process:Basically (11^10)/(10^10)=(1.1)^10=(1.1)^5*(1.1)^5, so we want powers of 1.1:1.11.211.3311.46411.61051 To get these powers of 1.1, simply take one row, shift it right, and add.Eg 1.1*1.1:1.1+0.11=1.21Eg 1.4641*111.4641+0.146410=1.61051Here we take approximation 1.1^5=1.6Now 1.1^5*1.1^5=1.6*1.6.In computing terms, this is a very common calculation as 16*16=256.So 1.1^10 = 1.1^5*1.1^5 = 1.6*1.6 = 2.56.Now you can try the same process for (11^20)/(10^20) and see how fast and simple this is !!!!
How to calculate X^Y ( X raised to the power y) by a basic/simple/financial calculator?
If y is an integer, you could decompose y into binary, and save some multiplications. Example: 1.73^34 34 = 100010 binary so 2^5 + 2^1 I'm going to assume you have an x^2 button, but if not, you might have to enter "*" then "=" 1.73 x^2 x^2 x^2 x^2 x^2 now the result is 1.73^32, save it +M 1.73 x^2 now we have 1.73^2, multiply by the saved 1.73^32 * MR = and you should have 1.73^34 Note that a basic calculator may run out of digits before long. The financial calculator may have a key for calculating Net Present Value or Interest. An interest calculation is nothing more than P*(1 + rate/100)^n so you could set your Principal to be 1, and choose the rate to make the right number in parentheses. Again, only if n is an integer. If you have access to the internet, you can just type "1.73^2.91 =" into Google, and will get the answer.
e to any power can be approximated efficiently with the formula:[math]e^x = \frac{x^0}{0!} + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}\dots[/math].To calculate [math]e^{-1.5}[/math]:[math]e^{-1.5} = \frac{-1/5^0}{0!} + \frac{-1.5^1}{1!} + \frac{-1.5^2}{2!} + \frac{-1.5^3}{3!}+ \frac{-1.5^4}{4!} \dots[/math][math]e^{-1.5} = 1 -1.5 + \frac{2.25}{2} - \frac{3.375}{6}+ \frac{5.0625}{24} \dots[/math][math]e^{-1.5} = 1 -1.5 + 1.125 - 0.5625 + 0.2109375 \dots[/math][math]\approx 0.2734375[/math]This is a pretty bad approximation (Google Calculator gives [math]0.22313016014[/math]) but it can be improved by using more and more terms of the sequence.
How can I do to the power of 0.5 without a calculator?
Find the square root of the number 4096 using division method. Solution: Process: 1. Put a dot on the digit in the units place. 2. Put dots on the numbers leaving digit form right to left. 3. First take down two numbers from left to right of the number. in this case it is 40. 4. Consider the perfect square number which is very near to 40. Here it is 36. 5. Then write down its square root namely 6 as divisor and also in the place of quotient. 6. Find the product of 6 and 6 which is 36. So write down 36 below the number 40 of the dividend. 7. Subtract 36 form 40, write down the remainder so write down 36 below the number 40 of the dividend. 8. Now bring down the digits up to the next dot, they are 96, write 96 beside 4 you get 496. 9. Double the previous quotient namely 6 you get 2*6 = 12. 10. Divide the two digits of the number from left to right. So obtained in step 8 by 12, you get 4. 11. Write 4 by the side of 12 to make the divisor 124. 12. Put 4 by the side of 6 the quotient to get 64. 13. Multiply 124 by 4, the product is 496. 14. Subtracting 496 from 496, you get zero as the remainder.
It factually relies on far too many details that are missing.ExplanationI explain in other articles OM, as operational Manipulations, and how these are Resistants to an Operational Path [of Least Resistance]Had this question been less Vague about:What is the Composition of Zero?What is the Desired Composition of One?we can indeed resolve these alike:[(.5 - .5)OM(+/-)] ÷ [(.5 -.5)]=OM(+/-)But do note, these Details do arise differences which best represent as an OR When these are not Compatible with OM, Forex:[(7 - 6) ÷ [(.5 -.5)]=[14 OR 12]And although that seemingly does not tell we a value with x=1; y=0 any OR, When properly used against a Lever=x+y in an equation will inform you:14x+12y=[14 or 12](Lever)=(7x+7y-6x-6y)/(.5–.5)And in this example, let we say x=8;y=4 to keep our numbers even, we observe 112+48=160=[1/0]×12 in this VERY Specific Instance.Without the Specifics, well...let me put it this way:Zero glass is far more fragile than one glass, so you gotta handle it with very much care.
The key thing to realize is that [math]1-i[/math] closely related to the roots of unity.[math]z \equiv 1-i = \sqrt{2} \left(\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}\right)[/math][math]=\sqrt{2}\left(\cos \frac{\pi}{4} - i \sin\frac{\pi}{4}\right)[/math][math]= \sqrt{2} \exp\left( i\frac{\pi}{4}\right)[/math]So [math]\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}[/math] is one of the eighth roots of unity — meaning that raising it to the eighth power is [math]1[/math] or all the solutions to [math]z^8 = 1[/math].So completing the steps:[math]z^4 = \left(\sqrt{2}\right)^4 \left( \exp\left( i\frac{\pi}{4}\right)\right)^4 = 4 \exp( i \pi)= - 4 [/math]
Calculate pH after 0.10 mol NaOH added to 1L solution...?
A) kb of methylamine is 4.4E-4 (don't forget to provide kb/ka values, because the ones that I use might be different from the one that your problem uses). When NaOH is added to methylammonium chloride, it produces methylamine completely. NaOH + CH3NH3Cl ----> CH3NH2 + NaCl + H2O 0.50 M * 1 L = 0.50 moles CH3NH2 0.60 M * 1 L = 0.60 moles CH3NH3Cl When you add 0.10 moles NaOH, it will completely react wiith CH3NH3Cl to form CH3NH2, so: 0.50 moles + 0.10 moles = 0.60 moles CH3NH2 0.60 moles - 0.10 moles = 0.50 moles CH3NH3Cl pH = pKa + log[A-/HA] kb = 4.4E-4, ka = kw/kb = 1E-14/4.4E-4 = 2.27E-11. pKa = 10.64 pH = 10.64 + log[0.60/0.50] = 10.64 + 0.08 = 10.72 B) Same thing but opposite. pH = pKa + log[A-/HA] You now have 0.40 M CH3CH2 and 0.70 M CH3NH3Cl by using a similar method from (A). pH = 10.64 + log[0.40/0.70] = 10.64 - 0.24 = 10.40