Hi
I've got two equations and I want to work out the X value where they intersect.
\(\displaystyle y=exp(ax+bx^2)\)
\(\displaystyle y=exp((cx+log(e))\)
a, b, c and e are constants. Since the lines intersect at y, I've tried to do:
\(\displaystyle exp(ax+bx^2)=exp((cx+log(e))\)
\(\displaystyle ax+bx^2=cx+log(e)\)
\(\displaystyle ax+bx^2 +cx=log(e)\)
\(\displaystyle (ca)x+bx^2 =log(e)\)
\(\displaystyle (ca)x+bx^2  log(e) = 0\)
\(\displaystyle bx^2+(ca)x  log(e) = 0\)
This, I believe, can be solved by the quadratic formula. But when I use sample values of
\(\displaystyle a = 0.041,
b=0.010,
c= 0.205,
e=1.210\)
I get answers of 1.09 or 6.18. I've got a graph of the intersection between the two curves, and it should be at around x=15.2 (see attached graph)
I don't understand where this is going wrong. I realise the graph is logarithmic and doesn't pass through the x axis. Is that why this is going wrong  if so, why can't the quadratic equation be used this way? Or am I just handling the quadratic wrongly?
Thanks for your help
L
I've got two equations and I want to work out the X value where they intersect.
\(\displaystyle y=exp(ax+bx^2)\)
\(\displaystyle y=exp((cx+log(e))\)
a, b, c and e are constants. Since the lines intersect at y, I've tried to do:
\(\displaystyle exp(ax+bx^2)=exp((cx+log(e))\)
\(\displaystyle ax+bx^2=cx+log(e)\)
\(\displaystyle ax+bx^2 +cx=log(e)\)
\(\displaystyle (ca)x+bx^2 =log(e)\)
\(\displaystyle (ca)x+bx^2  log(e) = 0\)
\(\displaystyle bx^2+(ca)x  log(e) = 0\)
This, I believe, can be solved by the quadratic formula. But when I use sample values of
\(\displaystyle a = 0.041,
b=0.010,
c= 0.205,
e=1.210\)
I get answers of 1.09 or 6.18. I've got a graph of the intersection between the two curves, and it should be at around x=15.2 (see attached graph)
I don't understand where this is going wrong. I realise the graph is logarithmic and doesn't pass through the x axis. Is that why this is going wrong  if so, why can't the quadratic equation be used this way? Or am I just handling the quadratic wrongly?
Thanks for your help
L
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