Question

logx^(5x-14)=2

Answer 1

Is this a calculus problem, or algebra?

Answer 2

logarithm question

Answer 3

I get that much; what course is it from?

Answer 4

calculus i guess at my collage we call it collage maths

Answer 5

It is an interesting problem. Let me look at it for a second.....

Answer 6

ok thanks

Answer 7

\[\log (x ^{5x-14})=2 \implies x ^{5x-14}=100 \implies (5x-14) \ln(x)= \ln(100)\]

Answer 8

I really am sort of wandering around here..... So...\[\frac{\ln(x)}{\ln(100)}=\log _{100}(x)=\frac{1}{{5x-14}} \implies x=100^{\frac{1}{{5x-14}}}\]

Answer 9

Not very helpful.... This is kind of an unusual exponential function.

Answer 10

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 ..

Answer 11

x^2-5x+14=0

Answer 12

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...

Answer 13

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.

Answer 14

The big problem I see with this is that b^2-4ac < 0, which does not bode well for a real number solution.

Answer 15

Sorry, I did what I could. I'll look in later, to see if somebody smarter than me can figure this one out.