find the value of x in : 2^x=x^2
2007-01-29 05:44:14 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
2^x = x^2
Taking log of both side with base 10
log10(2^x) = log10(x^2)
x*log10(2) = 2*log10(x)
By symmetry it is clear that x = 2 and 4 are two solutions.
2007-01-29 05:51:04 · answer #1 · answered by psbhowmick 6 · 2⤊ 0⤋
2
2007-01-29 05:52:09 · answer #2 · answered by MaryCheneysAccessory 6 · 0⤊ 0⤋
2
2007-01-29 05:48:55 · answer #3 · answered by bequalming 5 · 0⤊ 0⤋
2 and 4
2007-01-29 05:54:19 · answer #4 · answered by Anonymous · 0⤊ 0⤋
x = 2.
2007-01-29 05:48:50 · answer #5 · answered by Grizzly B 3 · 0⤊ 0⤋
If x=2 then the equation becomes 2^2=2^2.
2007-01-29 05:49:56 · answer #6 · answered by bruinfan 7 · 0⤊ 0⤋
You can see that x= 4 is a solution and also 2 but unless you use "Calculus" you can´t prove that there are no other solutions
2007-01-29 06:07:29 · answer #7 · answered by santmann2002 7 · 0⤊ 0⤋