SOLUTION: Please help me solve this equation: Solve the logarithmic equation logarithmic equation {{{ log( 2, (ln x) )= -1 }}}. <-- logarithm of lnx(natural log of x) with base 2 equals

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Please help me solve this equation: Solve the logarithmic equation logarithmic equation {{{ log( 2, (ln x) )= -1 }}}. <-- logarithm of lnx(natural log of x) with base 2 equals      Log On


   

Question 309530: Please help me solve this equation:
Solve the logarithmic equation logarithmic equation +log%28+2%2C+%28ln+x%29+%29=+-1+.
<-- logarithm of lnx(natural log of x) with base 2 equals =-1.
I tried:
+log%28+2%2C+%28log%28+e%2C+x+%29%29+%29=+-1+.
(log of natural log of x with base 2 equals -1)
+log%28+e%2C+x+%29=+2%5E-1+.
(natural log of x equals 2 to the power of -1)
+log%28+e%2C+x+%29=+%281%2F2%29+.
(natural log of x equals 1/2)
+x=+e%5E%281%2F2%29+.
(x equals e to the power of 1/2)
I put +e%5E%281%2F2%29+. back into the original equation and it didn't equal the answer.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Well +x=e%5E%281%2F2%29%5E%22%22+ is the answer.


+log%28+2%2C+%28ln+%28x%29%29+%29=+-1+ Start with the given equation.


+log%28+2%2C+%28ln+%28e%5E%281%2F2%29%5E%22%22%29%29+%29=+-1+ Plug in +x=e%5E%281%2F2%29%5E%22%22+


+log%28+2%2C+%28%281%2F2%29ln+%28e%29%29+%29=+-1+ Use the identity ln%28x%5Ey%29=y%2Aln%28x%29%29 to pull down the exponent.


+log%28+2%2C+%28%281%2F2%29%2A1%29+%29=+-1+ Take the natural log of 'e' to get 1.


+log%28+2%2C+%281%2F2%29+%29=+-1+ Multiply and simplify


+log%28+2%2C+%282%5E%28-1%29%29+%29=+-1+ Rewrite 1%2F2 as 2%5E%28-1%29


+-1%2Alog%28+2%2C+%282%29%29+%29=+-1+ Pull down the exponent using the identity log%28b%2C%28x%5Ey%29%29=y%2Alog%28b%2C%28x%29%29


+-1%2A1=+-1+ Evaluate the log base 2 of 2 to get 1.


+-1=+-1+ Multiply


Since the final equation is an identity (ie is always true), this verifies the answer.