Solving ln(ln(x)) = log(log(x))

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[@DarinBrownSJDCMath]

Looks like an identity to me...

[@mikeschieffer2644]

I got a different form of the value of x. x = e^(e^(ln(10)/(1-ln(10)))) which is approximately equal to 1.6940793. Using change of base formula I switched all of the log base 10 to natural logs. I ended up with the equation ln(ln(x)) = (ln(ln10))/(1-ln10). Then I used the exponential function twice to solve for x.

[@RR-bs9mr]

I also pretty sure there like a infinite number of imaginary solutions

[@josepherhardt164]

That's the way I did it. I'd forgotten the change-of-base formula and had to re-derive it first! Memories! It's actually rather remarkable that there is a non-imaginary answer to this problem.

[@Skank_and_Gutterboy]

That's what I did, too! I changed everything over to natural logs with the change of base formula.

[@anastasissfyrides2919]

I did the same.The thing is, i tried to plug in the solution just to verify and it doesn't make sense. Either i keep making arithmetic mistakes or we did smth wrong in the solution

[@Skank_and_Gutterboy]

@@anastasissfyrides2919 x=e^(ln10)^(1/(1-ln10)) checks good in MathCAD. The top post has an error in his solution, he's showing "e^(e^(ln(10)/(1-ln(10)))), he needs to remove the bottom "e^" and slide everything down. He took the correct solution and accidentally put it to the e-power. No!

[@robot8324]

Cool😃

[@farhansadik5423]

interesting as always!

[@SyberMath]

Glad you think so!

[@ManjulaMathew-wb3zn]

Halfway through the calculation it seems like x=1 is a solution but it makes the original equation undefined so had to be rejected. The domain of x is x>1 and the answer you got is within the domain of x>1

[@Skank_and_Gutterboy]

I got an answer that's a little more simplified: x=e^(ln10)^(1/(1-ln10)).

[@lawrencejelsma8118]

You forget to teach from scientists using ln(t) vs log(t) reasoning. If you are just creating a math formula the natural logarithm is the preferred choice over base ten in log(x). For scientific data where there is significantly large values of t on the t axis then log base 10 is preferred because base e the overall graph is bigger and consumes the page probably for some larger scientific data. If we just have a graph up to about 4 time constants of t then ln(t) allows a graph to magnify into the data graphed and scientists will plot on semilog charts in base e for up to about four time constants of t data. When t is checked for larger t (large time of the lower frequencies) semi log graphs of base 10 are more useful. Examples of that are Control Systems Transfer Functions defined in Bode plots!! 👍

[@mtaur4113]

Could simplify a bit more, reciprocal is the negative exponent, then multiply the powers.

[@mtaur4113]

Wolfram expresses the fractional power as a root, which looks sort of funny on the display but it checks out.

[@user-zy5tt8nu9h]

You forgot to say that lnx and logx >0, it means x>1

[@SyberMath]

You're right!

[@user-zy5tt8nu9h]

Don't worry, be happy😊

[@rakenzarnsworld2]

x = 1

[@duccline]

no, log(log1) = log(0) = undefined

[@jeanpaullamont]

ln(ln) or log only defined for x>1