0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

Step by step, from inside out:

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

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

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

$\endgroup$
0
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.