SOLUTION: What is X in (3x)^(1+ log of X base 3)=3 and how do I get to it?
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Question 1003768: What is X in (3x)^(1+ log of X base 3)=3 and how do I get to it?
Found 3 solutions by Alan3354, rothauserc, Theo:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
(3x)^(1+ log of X base 3)=3
---------
x = 1/9
-------------
Check:
(1/3)^(1-2) = 3
It's correct.
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
(3x)^(1 + log of x base 3) = 3
note that (3x)^(1 + log of x base 3) = 3 * x^(2 + (log x / log 3)
note that log is natural logarithm
3 * x^(2 + (log x / log 3) = 3
divide both sides of = by 3
x^(2 + (log x / log 3) = 1
take natural log of both sides of =
log(x)(2 + (log x / log 3) = 0
now write left side of = as a fraction with common denominator log(3)
log(x)(2log(3)+log(x)) / log(3) = 0
multiply both sides of = by log(3)
log(x)(2log(3)+log(x)) = 0
there are two solutions
1) log(x) = 0
x = 1
2) 2log(3) + log(x) = 0
log(x) = -2log(3)
note -2log(3) = log(1/(3^2)) = log(1/9)
x = 1/9
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
as far as i can tell, x = 1
when x = 1, the equation of:
(3x)^(1 + log3(x)) = 3 becomes:
(3*1)^(1 + log3(1)) = 3 which becomes:
(3)^(1 + log3(1)) = 3
log3(1) = y if and only if 3^y = 1
this can only occur if y = 0 because 3^0 = 1
therefore 1 + log3(1)) is equal to 1 + 0 which is equal to 1.
the equation becomes:
(3)^1 = 3 which becomes 3 = 3
the equation is true when x = 1.
how was this derived?
by magic of course.
here's the magic, if i did it correctly.
your equation is:
(3x)^(1+log3(x)) = 3
let y = 1 + log3(x) and your equation becomes:
(3x)^y = 3
this is true if and only if log3x(3) = y
log3x(3) is equal to log(3) / log(3x)
equation becomes log(3) / log(3x) = y
we had already set y = 1 + log3(x)
log3(x) is equal to log(x)/log(3)
therefore:
y = 1 + log3(x) becomes:
y = 1 + log(x)/log(3)
now we have:
y = 1 + log(x) / log(3) and we have:
y = log(3) / log(3x)
y = 1 + log(x) / log(3) is equivalent to y = (log(3) + log(x)) / log(3)
y = log(3) / log(3x) is equivalent to y = log(3) / (log(3) + log(x))
replace y in the second equation with y = (log(3) + log(x)) / log(3) from the first equation to get:
(log(3) + log(x)) / log(3) = log(3) / (log(3) + log(x)).
multiply both sides of this equation by log(3) and multiply both sides of this equation by (log(3) + log(x)) to get:
(log(3) + log(x))^2 = log(3)^2
take the square root of both sides of this equation to get:
log(3) + log(x) = log(3)
subtract log(3) from both sides of this equation to get:
log(x) = 0
solve for x to get x = 1.
there may be another, easier way to solve this, but i couldn't figure it out.
this was quite convoluted and i'm happy i was able to get an answer that made sense the way that i did do it.
hopefully you can understand what i did and why i did it.
some concepts involved.
y = logb(x) if and only if b^y = x
b^y = x if and only if logb(x) = y
log(a*b) = log(a) + log(b)
log(a^b) = b*log(a)
logb(x) = loga(x) / loga(b)
a + b/c = (ac + b) / c
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