# resolve: b) ln(ln(ln(lnx)))=0

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### 2 Answers

In logarithm we know that if;

lnx = y then;

`x = e^y`

In the same manner;

`ln(ln(ln(lnx))) = 0`

`(ln(ln(lnx))) = e^0 = 1`

`(ln(lnx)) = e^1 = e1`

`lnx = e^(e1)`

`x = e^(e^(e1))`

** So the answer is **`x = e^(e^(e1))`

**Sources:**

Solve ln(ln(ln(ln(x))))=0:

ln(ln(ln(ln(x))))=0 ==> ln(ln(ln(x)))=1 since ln(a)=0 ==>a=1; here a=ln(ln(ln(x)))

Now ln(ln(ln(x)))=1==>ln(ln(x))=e since ln(a)=1 ==>a=e

Then `ln(x)=e^e`

**And finally `x=e^(e^e)` **

Check:

`ln(x)=ln(e^(e^e))=e^e`

`ln(ln(x))=ln(e^e)=e`

`ln(ln(ln(x)))=ln(e)=1`

`ln(ln(ln(ln(x))))=ln(1)=0`

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