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The problem is: .5^x = 8^(2-x) The answer is: x=3 But how do you get that answer? I've been trying to figure it out for hours.... Someone please help....
ok i guess i cant do what i doe on facebook where i hit shift and enter at the same time so try to bare with me as i show my work.... : x log(.5)= (2-x)log(8) distribute: x log(.5)= 2log(8)- x log(8) get both x's on the same side: xlog(.5)+xlog(8)=2log(8) factor out the x: x(log[.5]+log[8])=2log8 divide both sides by (log[.5]+log[8]): x= 2log8/(log[.5]+log[8]) simply: x=3..... yeah its a lot of steps.... thanks for showing me the other way of doing it i think it might be easier...
Ok first you change your bases so they are the same and we can solve for the exponents. What's a base that can be used for both 0.5 and 8? Answer is 2.
So you get
2^(-1)(x)= 2(3)(2-x)
We didn't actually change the equation in any way because 2^(-1) is the equivalent of 0.5 and 2^3 is the equivalent of 8.
Now we can solve for the exponents. -1(x) = 3(2-x) -1x= 6 -3x 2x=6 x=3 Viola.
ok that seems easier then how my teacher explained it... he had me do it this way...
ok that seems easier then how my teacher explained it... he had me do it this way...
ok that seems easier then how my teacher explained it... he had me do it this way...
** 2^(-1)(x)=2^(3)(2-x)
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