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標題:

4^x+6^x =9^x, find x

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發問:

4^x+6^x =9^x, find x. Please help me

最佳解答:

4^x＋6^x＝9^x As 6^x＞0 for real x, we can divide the equation by 6^x and get (4/6)^x＋1＝(9/6)^x (2/3)^x＋1＝(3/2)^x Let y＝(2/3)^x, we get y＋1＝1/y y^2＋y＝1 y^2＋y－1＝0 y＝(±√5－1)/2 As y＝(2/3)^x＞0 for real x, we get y＝(√5－1)/2 As y＝(2/3)^x, we get ln y ＝x ln (2/3) x ＝ln y / ln (2/3)＝ ln [(√5－1)/2] / ln (2/3)＝1.1868

其他解答:

If you plot 4^x+6^x-9^x against x, you will notice a root at x=1.1865 (approx.). Zoomed in image at: http://i263.photobucket.com/albums/ii157/mathmate/4x_6x-1.png Using this information, you can refine the solution by rewriting the equation as xlog9=log(4^x+6^x) x=log(4^x+6^x)/log9 Starting with x=1.187 on the right hand side, and repeatedly substituting the new value of x on the RHS, we get successively: x1=1.186952665878328 x2=1.186917402828997 x3=1.186891132532082 Using Shank's transformation on x1,x2 and x3, we obtain x=(x1*x3-x2^2)/(x1+x3-2*x2)=1.186814389796409 accurate to 8 places of decimal. The accurate solution is: 1.186814390280981