The solution to (2^x)(e^(3x+1))=10 is (p + ln q)/(r+ln s), where all of p, q, r and s are integers. D=p+q+r+s.
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Originally Posted by roady The solution to $(2^x)(e^{3x+1})=10$ is (p + ln q)/(r+ln s), where all of p, q, r and s are integers. D=p+q+r+s. The solution of $(2^x)(e^{3x+1})=10$ for $x$ is $x\log(2)+(3x+1)=\log(10)\\x=\dfrac{\log(10)-1}{\log(2)+3}$
Originally Posted by roady The solution to (2^x)(e^(3x+1))=10 is (p + ln q)/(r+ln s), where all of p, q, r and s are integers. D=p+q+r+s. What is D in this context? I'm at a loss. -Dan
D is just the sum of the 4 integers p,q,r,s So here what I need to know is p=? q=? r=? s=?
Originally Posted by roady D is just the sum of the 4 integers p,q,r,s So here what I need to know is p=? q=? r=? s=? READ reply #2. Have you?
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