Solve For x

[nycmathguy]

Solve for x.

x^(x)^(5) = 100

 

[MathLover1]

do you have

or

 

[nycmathguy]

do you have

or

It's x raised to the x raised to the 5. It's the first choice.

 

[MathLover1]

where did you find this problem? you did not find this in precalculus

 

[nycmathguy]

where did you find this problem? you did not find this in precalculus

I found this problem in a FB math group. There are so many FB math groups. I don't recall where this is posted. You can disregard the question if it is beyond precalculus. Take a look at the recently-posted word problems in the Basic Math forum.

 

[MathLover1]

this requires more knowledge beyond calculus III , advanced Calculus

x^x^5 = 100......take log of both sides

Rewrite the equation with
2ln (10 )/5x=u
and
x= 2ln(10)/u5

(2ln(10)/u5 )e^-u=1 ..........rewrite in Lambert form

e^u*u=2ln(10)/5..........solve for u

u=W[0](2ln(10)/5)...............substitute back u=2ln(10)/5x

2ln(10)/5x=W[0](2ln(10)/5).........solve for x

2ln(10)/W[0](2ln(10)/5)=5x

x=2ln(10)/(W[0](2ln(10)/5)*5)

x = 10^(1/5)->solution


or, simplest way:
Graph each side of the equation. The solution is the x-value of the point of intersection.
x≈1.58489319

 

[nycmathguy]

this requires more knowledge beyond calculus III , advanced Calculus

x^x^5 = 100......take log of both sides

Rewrite the equation with
2ln (10 )/5x=u
and
x= 2ln(10)/u5

(2ln(10)/u5 )e^-u=1 ..........rewrite in Lambert form

e^u*u=2ln(10)/5..........solve for u

u=W[0](2ln(10)/5)...............substitute back u=2ln(10)/5x

2ln(10)/5x=W[0](2ln(10)/5).........solve for x

2ln(10)/W[0](2ln(10)/5)=5x

x=2ln(10)/(W[0](2ln(10)/5)*5)

x = 10^(1/5)->solution


or, simplest way:
Graph each side of the equation. The solution is the x-value of the point of intersection.
x≈1.58489319

Wow! I saw the beginning video clip and quickly thought it was a simple algebra 2 problem. Considering what you said, this problem belongs in a different forum.

 

[nycmathguy]

do you have

or

Here is the solution:

 

[MathLover1]

even simpler solution, I like it
better than use Lambert form

 

[nycmathguy]

even simpler solution, I like it
better than use Lambert form

Looks pretty easy to me.