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) About Me  (Show Source):
You can put this solution on YOUR website!
(3x)^(1+ log of X base 3)=3
---------
%283x%29%5E%281%2B+log%283%2CX%29%29=3
3x%2A%283x%29%5Elog%283%2CX%29%29=3
x%2A%283x%29%5Elog%283%2CX%29%29=1
x%2A%283%5Elog%283%2CX%29%29%2A%28x%5Elog%283%2Cx%29%29+=+1
x%2Ax%2A%28x%5Elog%283%2Cx%29%29+=+1
x%5E%28log%283%2Cx%29%29+=+1%2Fx%5E2+=+x%5E-2
log%283%2Cx%29+=+-2
x+=+3%5E%28-2%29
x = 1/9
-------------
Check:
(1/3)^(1-2) = 3
It's correct.


Answer by rothauserc(4718) About Me  (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) About Me  (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