SOLUTION: How can I find "x" when it's in this question? sqrt(log(x)-3) = log(x)-3 Also with this one... {{{e^(3ln(x^2)-2ln(x)) = ln(e^16)}}}

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: How can I find "x" when it's in this question? sqrt(log(x)-3) = log(x)-3 Also with this one... {{{e^(3ln(x^2)-2ln(x)) = ln(e^16)}}}      Log On


   

Question 369024: How can I find "x" when it's in this question?
sqrt(log(x)-3) = log(x)-3
Also with this one...
e%5E%283ln%28x%5E2%29-2ln%28x%29%29+=+ln%28e%5E16%29
Answer by CharlesG2(834) About Me  (Show Source):
You can put this solution on YOUR website!
How can I find "x" when it's in this question?
sqrt(log(x)-3) = log(x)-3
Also with this one...
e%5E%283ln%28x%5E2%29-2ln%28x%29%29+=+ln%28e%5E16%29

log is assumed to be log10 (log base 10)
sqrt(log(x) - 3) = log(x) - 3
sqrt(log(x) - log(1000)) = log(x) - log(1000) (log(1000) = 3)
logarithmic rule: log(m/n) = log(m) - log(n)
sqrt(log(x/1000)) = log(x/1000)
log(x/1000) = log(x/1000) * log(x/1000)
1 = log(x/1000) (divided by log(x/1000))
logarithmic rule: if log(y) = x, then b^x = y
10^1 = x/1000
10 = x/1000
10 * 1000 = x
10000 = x
check:
sqrt(log(10000) - 3) = log(10000) - 3
sqrt(4 - 3) = 4 - 3
sqrt(1) = 1
1 = 1, yes

e%5E%283ln%28x%5E2%29-2ln%28x%29%29+=+ln%28e%5E16%29
logarithmic rule: nln(m) = ln(m^n)
e%5E%28ln%28%28x%5E2%29%5E3%29+-+ln%28x%5E2%29%29+=+16
e%5E%28ln%28x%5E6%29+-+ln%28x%5E2%29%29+=+16
e%5E%28ln%28x%5E6%2Fx%5E2%29%29+=+16
e%5E%28ln%28x%5E4%29%29+=+16
logarithmic rule: if b^x = y, then ln(y) = x
ln%2816%29+=+ln%28x%5E4%29
what is in parenthesis needs to equal
16+=+x%5E4
2+=+x
check:
e^(3ln(2^2)-2ln(2)) = ln(e^16)
e^(3ln(4) - 2ln(2)) = 16
e^(ln(4^3) - ln(2^2)) = 16
e^(ln(64) - ln(4)) = 16
e^(ln(64/4)) = 16
e^(ln(16)) = 16
ln(16) = ln(16), yes