anonymous
  • anonymous
logx^(5x-14)=2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Is this a calculus problem, or algebra?
anonymous
  • anonymous
logarithm question
anonymous
  • anonymous
I get that much; what course is it from?

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anonymous
  • anonymous
calculus i guess at my collage we call it collage maths
anonymous
  • anonymous
It is an interesting problem. Let me look at it for a second.....
anonymous
  • anonymous
ok thanks
anonymous
  • anonymous
\[\log (x ^{5x-14})=2 \implies x ^{5x-14}=100 \implies (5x-14) \ln(x)= \ln(100)\]
anonymous
  • anonymous
I really am sort of wandering around here..... So...\[\frac{\ln(x)}{\ln(100)}=\log _{100}(x)=\frac{1}{{5x-14}} \implies x=100^{\frac{1}{{5x-14}}}\]
anonymous
  • anonymous
Not very helpful.... This is kind of an unusual exponential function.
anonymous
  • anonymous
i know, i was kinda thinking i could work it out back to a simultaneous equation by making the 2 to the power of x ..
anonymous
  • anonymous
x^2-5x+14=0
anonymous
  • anonymous
OK, so it goes like this......\[\log(x ^{5x-14})=2 \implies x ^{5x-14}=100 \implies (5x-14)=\log_x(100) \implies \]This is just different forms of the same stuff I had before. I don't see where your polynomial comes from...
anonymous
  • anonymous
Maybe I misunderstood the problem. Did it start out like this?\[\log_x(5x-14)=2\]If so, your polynomial is correct. If the problem is as stated at the top of this column, I don't see any way it comes to that.
anonymous
  • anonymous
The big problem I see with this is that b^2-4ac < 0, which does not bode well for a real number solution.
anonymous
  • anonymous
Sorry, I did what I could. I'll look in later, to see if somebody smarter than me can figure this one out.

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