I can't come up with any way to solve it algebraically, and KAUSH's way doesn't work, after log3 x = 1/x you change that to 3^(1/x) = x , which is the same thing you'd get if you just took the xth root of both sides from the beginning.
The best I can do is use trial and error to get roughly x = 1.82545
Hope this helps.
2006-07-28 19:56:09
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answer #1
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answered by CubicMoo 2
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a million/3(x + 6)= 11 Multiply the two aspects via 3. (a million/3)(3) cancel one yet another out (11)(3) = 33 (x+ 6)= 33 Subtract 6 from the two aspects. 6-6=0.33-6=27 So x=27 to verify this do 27+6 which equals 33 than multiply that via a million/3 which provides 11. 11=11 So that's right!! desire I helped!!
2016-11-03 06:02:29
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answer #2
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answered by ? 4
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1.82545502292483
There is no normal way to solve this problem. But there are several ways to calculate the answer through trial and error methods. One way is to graph y = x^x on a calculator, and then find the value of x where y = 3.
Another way is using a computer program, which is what I've done.
Here is the java-script.
You can save that text in a .js file and run it from any windows computer.
max_x = 4
min_x = 0
x = 2
for(loop = 0; loop < 1000; loop++)
{
y = Math.pow(x,x);
if (y > 3) max_x = x;
if (y < 3) min_x = x;
x = (max_x + min_x) / 2.0;
}
WshShell = WScript.createObject("WScript.Shell");
WshShell.Popup(x);
2006-07-28 20:23:01
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answer #3
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answered by Michael M 6
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Using base 10 logs (so you can use log tables) you get log(x)=log(3)/x; log(3)=.477, so the equation is
log(x) = .477/x
This is a transcendental equation and can be solved graphically or by successive approximations. The result is x=1.825
2006-07-28 20:10:24
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answer #4
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answered by gp4rts 7
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x^x=3
taking log on both sides with base 3
x log3 X = log3 3
x log3 X = 1
log3 X= 1/x
now you can solve it yourself
2006-07-28 19:49:12
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answer #5
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answered by Anonymous
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It needs to be done by trial & error. And the correct answer is in between 1.82 & 1.83.
2006-07-28 21:09:06
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answer #6
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answered by sharanan 2
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x2=3
square root 3=x
2006-07-28 19:38:04
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answer #7
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answered by al 2
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