How to find "y" from this equation (involve ln and e)

y = e^(1/2 ln2) + 2e^[-(1/2 ln2)]

the answer is 2(*squareroot*2)
or we can say as *squareroot*8

Steps please. Thank you

Wow man this is a serious problem.

The is a rule

Rule:

xln(z) = ln(z^x).

lets break into 2 parts.

Part 1:

e^(1/2ln(2)) = e^(ln(2^1/2))

Well we are bummed but there is another rule called the COLOR OF THE WHITEHOUSE!

Rule:

e^(ln(2)) = 2
10^(log(88)) = 88

so using the rule above we get e^(ln(2^(1/2))) = 2^(1/2)

Part 2:

2e^[-(1/2ln(2))] = 2e^[(ln(2^(-1/2))] = 2/(2^(1/2)) = 2^(1/2)

now we add!

2^(1/2) + 2^(1/2) = 2sqrt(2) or 8^(1/2) sqaure root of 8!!!!!!!!!!!






COLOR OF THE WHITEHOUSE

Answer :-

2.8284;

Explanation :-

Use the following identity;

x = e^(lnx) --- --- --- (1);

let,

x = 2;

and,

rx =square-root of x;

First term = e^(ln(rx)) = rx;

Second term = 2*e^(ln(1/rx))= 2/rx;

y = rx + 2/rx = rx+ rx = 2*rx = 2*1.414 = 2.8284

y = e^(1/2 ln2) + 2e^[-(1/2 ln2)]
= 2^(1/2) + 2/2^(1/2)
= (2 + 2 )/2^(1/2) * 2^(1/2)/2^(1/2)
= [4(2^1/2)]/2
= 2*2^(1/2)

Edit: yes sqrt8 = 2 * sqrt2