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can u please teach me how to solve exponents manually? thanks in advance
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What does "exponents" mean at all?
If it's a^x=b, then x=ln(b)/ln(a)
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please give some example.
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for example, 2^59 or 2^-59... without using any calculator.. in smaller numbers for example, 2^3=8...
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my question is how to get the n in 2^59=n or 2^-59=n... thanks again.
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I have a question.. Do you know what
means? If you do, I can't see how you can have problems calculating it manually.Support MathsIsFun.com by clicking on the banners.
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aha!
I don't know any method. (which gives exact)
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I have a question.. Do you know what
means? If you do, I can't see how you can have problems calculating it manually.
But he is talking about 2^59, which is much more hard to compute.
Here is it:
2^59=576460752303423488
2^-59=1/(2^59)≈ 1.734723475976807094411924481391906738281e-18
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Sorry Patrick.
You mean the definition of exponent?
If b is a positive integer integer, then
a^b=a.a.a.a. ... .a --->b times
For example a^1=a; a^2=a.a; a^3=a.a.a
By definition a^0=1;
If b is negative, we use:
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Patrick wrote:I have a question.. Do you know what
means? If you do, I can't see how you can have problems calculating it manually.But he is talking about 2^59, which is much more hard to compute.
Here is it:
2^59=576460752303423488
2^-59=1/(2^59)≈ 1.734723475976807094411924481391906738281e-18
yeah ur right... i can only calculate exponents manually if small values are involved.. Sorry, but i still can't understand how you came up with that answer... any other methods?
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or let us assume that i am not using a scientific calculator? how will i calculate the exponents?
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or let us assume that i am not using a scientific calculator? how will i calculate the exponents?
as long as it is a small integer values, you can just do
x^n = x*x*x*x.....n times
x^-n = 1/(x*x*x*x....n times)
if it is fractional, and it is a half, most non scientific calculates will still carry a square root
x^0.5 = √x
x^-0.5 = 1/√x
and from there, aslong as the exponent is 1/ some multiple of 2, you can use square roots, so example
x^0.25 = √√x
x^0.125 = √√√x
x^0.0625 = √√√√x
another important rule, is that x^(a/b) = (b root of x)^a
so for example x^3/4 = 4th root of x, cubed, so aslong as the exponent is some combination of 1/ some multiple of 2, you can get it with non scientific too, for example
x^0.75 = x^(3*0.25) = (√√x)*(√√x)*(√√x) = (√x)*(√√x) -> x^(0.5+0.25)
x^0.625 = x^(6*0.125) = (√√√x)*(√√√x)*(√√√x)*(√√√x)*(√√√x)*(√√√x) = (√x)*(√√√x) -> x^(0.5+0.125)
Last edited by luca-deltodesco (2006-07-23 02:41:22)
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what if, i only have a pen and a paper?
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For example, non-presice result for 2^59:
2^10=1024>10^3;
2^3=8<10;
With choosing different numbers, you may get more accurate results.
Last edited by krassi_holmz (2006-07-23 03:06:53)
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well. you cant.. not unless its obvious
like what is 2^3, 2*2=4, *2 = 8
or 4^0.5 = 2
27^(2/3) = 9 -> 27^1/3 = cube root, = 3, squared = 9
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well. you cant.. not unless its obvious
like what is 2^3, 2*2=4, *2 = 8
or 4^0.5 = 2
27^(2/3) = 9 -> 27^1/3 = cube root, = 3, squared = 9
Yes.
For example, 2^93=9903520314283042199192993792, which is very close to 10^...
But there's another problem-you can't practically use the above, because you must know fast way to evaluate 2^k for every positive integer k<93
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