SOLUTION: Solve. Give exact solution with approximate solution to four decimal places ln(ln x)=8

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Question 248891: Solve. Give exact solution with approximate solution to four decimal places

ln(ln x)=8
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
ln%28ln%28x%29%29+=+8
Solving equations where the variable is in the argument of a logarithm usually involves transforming the equation into one of the following forms:
  • log(variable-expression) = other=expression
  • log(variable-expression) = log(other-expression)

Your equation is already in the first form so we will go with that form. With this form we proceed by:
  1. Using the fact that log%28a%2C+%28p%29%29+=+q is equivalent to p+=+a%5Eq to rewrite the equation in exponential form.
  2. Solve the resulting equation.
  3. Check the solution(s). This is important with logarithmic equations. We must reject any solutions that make the argument to a any logarithm zero or negative.

So we start by
1) rewriting in exponential form. This gives us:
ln%28x%29+=+e%5E8 Next we
2) Solve the resulting equation. This is an equation with the variable in the argument of a logarithm (of the first form). So we repeat the procedure with this equation. Rewrite in exponential form:
x+=+e%5E%28e%5E8%29
This equation, fortunately, is already solved!
3) Check your solution(s). Always use the original equation to check your solutions:
ln%28ln%28x%29%29+=+8
Checking x+=+e%5E%28e%5E8%29:
ln%28ln%28e%5E%28e%5E8%29%29%29+=+8
Since e%5E%28e%5E8%29 is positive, the inner logarithm has a positive argument (which is what we want. And since e%5E%28e%5E8%29 is greater than 1, ln%28e%5E%28e%5E8%29%29 is also positive. So the outer logarithm will also be positive. We have no reasons to reject our solution. If we wish to complete the check then we can use some properties of logarithms to simplify the left side. One property of logarithms is
By definition ln(e) = 1 so this becomes:
ln%28e%5E8%2A1%29+=+8
ln%28e%5E8%29+=+8
Now we can repeat this for the remaining logarithm:
8%2Aln%28e%29+=+8
8%2A1+=+8
8+=+8 Check!

To get a decimal approximation for x+=+e%5E%28e%5E8%29 we start with a decimal approximation for e. e is approximately equal to 2.7182818284590451. (Note: You do not have to use all these debimal places. But I would not rould off to just four places until the end. If you have calculator software on your computer that can do exponents and you know how to copy and paste, then I would just copy this number for e and paste it into the calculator.)

So
x+=+e%5E%28e%5E8%29 becomes:
x+=+%282.7182818284590451%29%5E%282.7182818284590451%5E8%29
which is a very large number (approximately):
x+=+4.1078+%2A+10%5E1294

P.S. If your equation is (or can be transformed into) the second form:
log(variable-expression) = log(other-expression)
(where the bases of the logs are the same) you just set
variable-expression = other-expression
and solve that equation.