Asked by chandra
How do you solve (3^x)(3x^x+1)=9
Answers
Answered by
drwls
Solve it by iteration or graphing.
If x = 0, (3^x)(3x^x+1)=1*4 = 4
If x=1, (3^x)(3x^x+1)= 3*4 = 12
If x=0.5, (3^x)(3x^x+1)=(1.732)(3.121) = 5.406
If x = 0.7, (3^x)(3x^x+1)= (2.158)(3.208) = 6.923
If x=0.8, (3^x)(3x^x+1)= (2.408)(3.510)= 8.451
If x=0.9, (3^x)(3x^x+1)= (2.688)(3.729)= 10.02
Interpolating, the answer is approximately x = 0.835
For x=0.835,
(3^x)(3x^x+1)= (2.503)(3.581) = 8.962
If x = 0, (3^x)(3x^x+1)=1*4 = 4
If x=1, (3^x)(3x^x+1)= 3*4 = 12
If x=0.5, (3^x)(3x^x+1)=(1.732)(3.121) = 5.406
If x = 0.7, (3^x)(3x^x+1)= (2.158)(3.208) = 6.923
If x=0.8, (3^x)(3x^x+1)= (2.408)(3.510)= 8.451
If x=0.9, (3^x)(3x^x+1)= (2.688)(3.729)= 10.02
Interpolating, the answer is approximately x = 0.835
For x=0.835,
(3^x)(3x^x+1)= (2.503)(3.581) = 8.962
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