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Sarai Davenport

Sarai Davenport

Answered question

2022-06-18

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?

Answer & Explanation

EreneDreaceaw

EreneDreaceaw

Beginner2022-06-19Added 20 answers

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
( ln ( exp ( 32 ) ) ) 1 / 2
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.
George Bray

George Bray

Beginner2022-06-20Added 12 answers

Step by step, from inside out:
log ( e 64 ) 1 / 2 = 1 2 log e 64 = 1 2 64 = 32
[ log ( e 64 ) 1 / 2 ] 1 / 2 = 32 = 4 2
log ( e 4 2 ) 1 / 2 = 1 2 4 2 = 2 2 = 8

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