$\ln( \exp( \ln( \exp( 64 )^{1/2} )^{1/2} )^{1/2} )$

I keep getting the answer 8. But the textbook as well as wolfram say it's 8^(1/2) or in other words 2(2^(1/2)).

Here are the steps I took, basically just following the rules of logarithms.

$\ln( \exp( \ln( \exp( 64 )^{1/2} )^{1/2} )^{1/2} )$

$\ln( \exp( \ln( \exp( 32 ) )^{1/2} )^{1/2} )$

$\ln( \exp( (1/2)\ln( \exp(32) ) )^{1/2} )$

$\ln(\exp((1/4)\ln(\exp(32))))$

$\ln(\exp((8))$

$8$

I've worked through it several times and keep getting this answer. Am I missing something?

Step by step, from inside out:

$$\log \left(e^{64}\right)^{1/2}=\frac12\log e^{64}=\frac1264=32$$

$$\left[\log \left(e^{64}\right)^{1/2}\right]^{1/2}=\sqrt{32}=4\sqrt2$$

$$\log\left(e^{4\sqrt2}\right)^{1/2}=\frac124\sqrt2=2\sqrt2=\sqrt8$$

On your question you need a closing parenthesis after exp(64). Then you will have the answer.

From your second line to your third line, you bring an exponent of $1/2$ out in front of a logarithm, but not in a valid way.

$$\ln(\exp(32))^{1/2}$$

means the same as

$$\big(\ln(\exp(32))\big)^{1/2}$$

where pulling that $1/2$ out front is not legal. It does not mean the same as

$$\ln\big((\exp(32))^{1/2}\big)$$

where pulling the exponent down to the front would be valid. This is just a matter of what the standard order of operations dictates: function evaluation has higher precedence than exponentiation.