Question

I need a genius. This a a university entrance exam's question: Find x^-1 . This is algebra. A graphing calculator is not allowed.

Answer 1

hmm is that all the information given/

Answer 2

Here's the equation \[x^{x^{\frac{1}{2}}} = \frac{1}{\sqrt{2}}\] I need to find: \[x^{-1}\]

Answer 3

remember the exponential rule:
\[a^{-m}={1\over a^m}\]
therefore,
\[x^{-1}={1\over x}\]
for each number, take the reciprocal

Answer 4

@amistre64

Answer 5

@phi

Answer 6

@.Sam.

Answer 7

@jim_thompson5910

Answer 8

@terenzreignz

Answer 9

\[x^{x^{\frac{1}{2}}} = \frac{1}{\sqrt{2}}\]
\[log(x^{x^{\frac{1}{2}}}) = log(\frac{1}{\sqrt{2}})\]
\[x^{1/2}log(x) = log(1)-log(\sqrt{2})\]
\[x^{1/2}log(x) = -log(\sqrt{2})\]
hmmm, not sure if this is gonna pan out this way

Answer 10

To be honest I don't think log is needed since it's a question from Chapter 2. It's all about exponents.

Answer 11

x^(1/2) = -1 might be what they are asking for then

Answer 12

is that\[x^{(x^{1/2})}~or~(x^x)^{1/2}\]

Answer 13

Without parenthesis

Answer 14

They both have parentheses?

Answer 15

they are not the same ....
http://www.wolframalpha.com/input/?i=y%3D%28x%5Esqrt%28x%29%2C+y+%3D+sqrt%28x%5Ex%29

Answer 16

the first one.

Answer 17

the second one is totally different

Answer 18

then for the first one, id assume that x^(-1) = x^(sqrt(x)), when -1 = sqrt(x)

Answer 19

but thats not taking into account the ... = sqrt(2)/2 stuff

Answer 20

\[x^{x^{1/2}}={1\over\sqrt{2}}\\
\implies \sqrt{2}\times x^{x^{1/2}}=1\\
\implies x^{1/2}=0\\
\implies x=0
\]
the given equation cannot be satisfied

Answer 21

atleast not for \(x\in\mathcal{R}\)

Answer 22

How exactly does this follow?
\[x^{x^{1/2}}={1\over\sqrt{2}}\\
\implies \sqrt{2}\times x^{x^{1/2}}=1\\
\color{red}{\implies x^{1/2}=0}\\
\implies x=0\]