Finding the nth Exponent: Solving for n in 2^2^n = 36 trillion

  • Thread starter MrAnderson
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In summary, the conversation is about finding the value of n in the equation 2^2^n = 36 trillion. The solution involves taking the log on both sides and using some algebraic manipulation to solve for n. The final value of n is approximately 5.4929.
  • #1
MrAnderson
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Homework Statement


find n.



Homework Equations


2^2^n = 36trillion



The Attempt at a Solution


I have worked this out and found n to be equal to 3.67 can someone confirm that i have the right solution? I took the log on both sides and worked it out.
 
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  • #2
I get n ~ 5.4929. You have to take the log of both sides twice.
For my value of n, 2^n ~ 45.03, and 2 to that power is close to 36 trillion.
 
  • #3
Mark44 said:
I get n ~ 5.4929. You have to take the log of both sides twice.
For my value of n, 2^n ~ 45.03, and 2 to that power is close to 36 trillion.

That's correct that 2^n = 36trillion and n=45, i get this also. But when i attempt to get n when 2^2^n = 36 trillion i first convert 2^nlog2 =log(36trillon)
which becomes log 2^nlog2 =log(log(36 trillion)), now at this point i divide both sides by log2. Now I am totally stuck and lost. Can you explain the process?
 
  • #4
2^(2^n) = 36 x 10^12
==> (2^n) log 2 = log(36 x 10^12)
==> n log(2) + log(log 2) = log(log(36 x 10^12))

Now, subtract log(log 2) from both sides, and then divide both sides by log(2).

I think you did something weird with the exponents when you got 2^nlog2. The exponent on 2 of your original express is 2^n, so when you take the log of 2^2^n, you get 2^n * log 2, and not what you show.
 
  • #5
Mark44 said:
2^(2^n) = 36 x 10^12
==> (2^n) log 2 = log(36 x 10^12)
==> n log(2) + log(log 2) = log(log(36 x 10^12))

Now, subtract log(log 2) from both sides, and then divide both sides by log(2).

I think you did something weird with the exponents when you got 2^nlog2. The exponent on 2 of your original express is 2^n, so when you take the log of 2^2^n, you get 2^n * log 2, and not what you show.

Oh i see, you have to add them and then subtract them then divide by log(2). thanks a lot, you've been very helpfull :).
 

1. What is "nth exponent"?

The nth exponent is a mathematical notation used to represent the power to which a number is raised. It is commonly denoted as "n" and represents the position of the exponent in a sequence, starting from 0.

2. How do you solve for the nth exponent in an equation?

To solve for the nth exponent in an equation, you will need to use logarithms. Take the logarithm of both sides of the equation and then use the power rule to isolate the exponent. Finally, solve for "n" by equating the resulting logarithmic expression to the given value.

3. Can the nth exponent be a negative or a decimal number?

Yes, the nth exponent can be any real number, including negative and decimal numbers. This is because exponents represent repeated multiplication, and any number can be multiplied by itself any number of times, including non-whole numbers.

4. What are some real-life applications of solving for the nth exponent?

Solving for the nth exponent is useful in various fields such as finance, physics, and computer science. For example, it is used to calculate compound interest, determine radioactive decay, and analyze algorithms in computer programming.

5. Are there any special cases when solving for the nth exponent?

Yes, there are special cases when solving for the nth exponent, such as when the base is negative. In these cases, the value of "n" must be an odd integer for the solution to be real. Additionally, when the base is 0, the value of "n" must be positive for the equation to have a solution.

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