SOLUTION: This is the new one: log(25,1/5.log(3,2-log(0.5,x)))=-1/2

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Question 202723: This is the new one:
log(25,1/5.log(3,2-log(0.5,x)))=-1/2
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
If the equation is
log%2825%2C%28%281%2F5%29log%283%2C%282-log%280.5%2Cx%29%29%29%29%29=-1%2F2
then the solution, which I call "peel the onion", follows. If this equation is not correct, try again this time:
  • Start your equation with three left braces (shift [) and finish the equation with three right braces (shift ])
  • Use more parentheses to make sure everything is properly grouped
    Although not required, we are going to use an additional variable, z, to help make what we are doing easier to understand.

    First we'll make z+=+%281%2F5%29log%283%2C%282-log%280.5%2Cx%29%29%29
    Substituting z into the original equation we get:
    log%2825%2C+z%29+=+-1%2F2
    Rewriting this in exponential form we get:
    z+=+25%5E%28-1%2F2%29+=+1%2F%2825%5E%281%2F2%29%29+=+1%2F%28sqrt%2825%29%29+=+1%2F5
    So z = 1/5. Now we will substitute back the original expression for z:
    %281%2F5%29log%283%2C%282-log%280.5%2Cx%29%29%29+=+1%2F5
    Multiplying both sides by 5 we get:
    log%283%2C%282-log%280.5%2Cx%29%29%29+=+1
    Now we will reuse z (or use another variable if this make you uncomfortable). Let z = the argument of the outer log function.
    z+=+2-log%280.5%2Cx%29%29
    Substituting:
    log%283%2C+z%29+=+1
    Rewriting in exponential form:
    z+=+3%5E1+=+3
    Substituting back for z:
    2-log%280.5%2Cx%29+=+3
    Subtracting 2 from both sides we get
    -log%280.5%2C+x%29+=+1
    Dividing (or multiplying) both sides by -1 we get
    log%280.5%2C+x%29+=+-1
    Rewriting in exponential form we get
    x+=+%280.5%29%5E%28-1%29+=+%281%2F2%29%5E%28-1%29+=+2%2F1+=+2
    So our solution is: x = 2. I call this "peel the onion" because the x was buried inside a log which was buried inside anothe log which was buried inside a 3rd log. So I "peeled away" from the outside until x was by itself.