Question 474932: x^logx = 100x
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! your equation is:
x^(log(x)) = 100x
divide both sides of this equation by x to get:
(x^(log(x))/x = 100
take the log of both sides of this equation to get:
log(x^(log(x))/x) = log(100)
since log(a/b) = log(a) - log(b), your equation becomes:
log(x^(log(x))) - log(x) = log(100)
since log(a^b) = b*log(a), your equation becomes:
log(x) * log(x) - log(x) = log(100)
subtract log(100) from both sides of this equation to get:
log(x) * log(x) - log(x) - log(100) = 0
since log(100)= 2, your equation becomes:
log(x) * log(x) - log(x) - 2 = 0
simplify this to get:
log^2(x) - log(x) - 2 = 0
let y = log(x) and your equation becomes:
y^2 - y - 2 = 0
this is a quadratic equation that can be solved by factoring.
you get:
y^2 - y - 2 = (y-2) * (y+1)
solving for y gets you:
y = 2
or:
y = -1
since y = log(x), this means that:
log(x) = 2
or:
log(x) = -1
log(x) = 2 means that:
x = 100
log(x) = -1 means that:
x = .1
your answers are that:
x = 100 or x = .1
plug those into your original equation to see if those equations hold true.
your original equation is:
x^(log(x)) = 100x
replace x with 100 and the equation becomes:
100^(log(100)) = 100*100.
this becomes:
100^2 = 100*100 which becomes:
100^2 = 100^2 which is true, confirming the value of x = 100 is good.
replace x with .1 and the equation becomes:
.1^(log(.1)) = 100*.1
this becomes:
.1^(-1) = 100*.1
since a^-b = 1/a^b, this becomes:
1/(.1^1) = 100*.1 which becomes:
1/(.1) = 100*.1 which becomes:
10 = 10 which is true, confirming the value of x = .1 is good.
this was a tricky problem, but if you just follow the rules, you will arrive at the answer, as i did.
when you see a problem like a^b = c, you take the log of both sides of that equation and solve.
the fact that this had a log(x) as the exponent didn't change the nature of the problem.
recognizing that you had a quadratic type of problem also was critical to solving this problem.
a quadratic type of problem is in the form of ax^2 + bx + c = 0
since i already was using the letter x, i changed the standard form of the quadratic equation to ay^2 + by + c = 0
I then said that y = log(x) and the equation of:
log^2(x) - log(x) - 2 = 0 became:
y^2 - y - 2 = 0
if you compare this to the standard form of ay^2 + ay + c = 0, you will see that:
a = 1
b = -1
c = -2
i could have solved the quadratic equation by using the quadratic formula but didn't need to since i was able to factor that equation directly.
you also need to know the laws of logarithms as well as the laws of exponents.
become familiar with those and the problems becomes easier to solve.
any additional questions you have involving logarithms, just let me know and i'll help you out as much as i can.
once again, this was tricky, but solvable.
log(x) = 2 means that x = 100 is solved by using the basic law of logarithms that states:
y = log(b,x) if and only if b^y = x
log(x) = 2 is equivalent to:
log(10,x) = 2
the base of the logarithm is 10.
by the basic law of logarithms:
log(10,x) = 2 if and only if 10^2 = x
this means that 100 = x which means that x = 100.
a similar process was used to determine that log(10,x) = -1 if and only if 10^(-1) = x which made x = .1.
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