Question 202723
If the equation is
{{{log(25,((1/5)log(3,(2-log(0.5,x)))))=-1/2}}}
then the solution, which I call "peel the onion", follows. If this equation is not correct, try again this time:<ul><li>Start your equation with three left braces (shift [) and finish the equation with three right braces (shift ])</li><li>Use more parentheses to make sure everything is properly grouped</li</ul>
Although not required, we are going to use an additional variable, z, to help make what we are doing easier to understand.<br>
First we'll make {{{z = (1/5)log(3,(2-log(0.5,x)))}}}
Substituting z into the original equation we get:
{{{log(25, z) = -1/2}}}
Rewriting this in exponential form we get:
{{{z = 25^(-1/2) = 1/(25^(1/2)) = 1/(sqrt(25)) = 1/5}}}
So z = 1/5. Now we will substitute back the original expression for z:
{{{(1/5)log(3,(2-log(0.5,x))) = 1/5}}}
Multiplying both sides by 5 we get:
{{{log(3,(2-log(0.5,x))) = 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(0.5,x))}}}
Substituting:
{{{log(3, z) = 1}}}
Rewriting in exponential form:
{{{z = 3^1 = 3}}}
Substituting back for z:
{{{2-log(0.5,x) = 3}}}
Subtracting 2 from both sides we get
{{{-log(0.5, x) = 1}}}
Dividing (or multiplying) both sides by -1 we get
{{{log(0.5, x) = -1}}}
Rewriting in exponential form we get
{{{x = (0.5)^(-1) = (1/2)^(-1) = 2/1 = 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.