# How do you solve  log _(2) (log _(2) (log _(2) x))) = 2?

Dec 17, 2015

Make repeated use of the fact that ${2}^{{\log}_{2} \left(x\right)} = x$ to find that
$x = 65536$

#### Explanation:

Applying the property of logarithms that

${a}^{{\log}_{a} \left(x\right)} = x$

we have

${\log}_{2} \left({\log}_{2} \left({\log}_{2} \left(x\right)\right)\right) = 2$

$= {2}^{{\log}_{2} \left({\log}_{2} \left({\log}_{2} \left(x\right)\right)\right)} = {2}^{2}$

$\implies {\log}_{2} \left({\log}_{2} \left(x\right)\right) = 4$

$\implies {2}^{{\log}_{2} \left({\log}_{2} \left(x\right)\right)} = {2}^{4}$

$\implies {\log}_{2} \left(x\right) = 16$

$\implies {2}^{{\log}_{2} \left(x\right)} = {2}^{16}$

$\implies x = 65536$

Dec 17, 2015

$x = {2}^{16} = 65536$

#### Explanation:

Start from this point: if you know that $a = b$, then it must be ${2}^{a} = {2}^{b}$. So, start from your equation, and deduce that

${2}^{{\log}_{2} \left({\log}_{2} \left({\log}_{2} \left(x\right)\right)\right)} = {2}^{2}$

Now, by definition, ${2}^{{\log}_{2} \left(z\right)} = z$, while of course ${2}^{2} = 4$. So, the equation becomes

${\log}_{2} \left({\log}_{2} \left(x\right)\right) = 4$

And now we're in the same situation as before, so we can apply the same logic:

${2}^{{\log}_{2} \left({\log}_{2} \left(x\right)\right)} = {2}^{4}$

Which means

${\log}_{2} \left(x\right) = 16$

One last iteration gives us

$x = {2}^{16} = 65536$

${\log}_{2} \left({\log}_{2} \left({\log}_{2} \left({2}^{16}\right)\right)\right) = {\log}_{2} \left({\log}_{2} \left(16\right)\right) = {\log}_{2} \left(4\right) = 2$