If {sqrt(3)+sqrt(2)}^x + {sqrt(3)-sqrt(2)}^x =10 then find the value of x ?

2 Answers
Apr 9, 2018

pm2

Explanation:

Calling

a = sqrt3+sqrt2
b = sqrt3-sqrt2

we have the equivalent problem

{(a^x+b^x=10),(ab=1):}

or

a^x+a^-x = 10 or

e^(lambda x)+e^(-lambda x) = 10 or

2cosh(lambda x) = 10 or

x = 1/lambda "arccosh"(5)

Here lambda = lna=ln(sqrt3+sqrt2) so finally

x = ("arccosh"(5)) /ln(sqrt3+sqrt2) =2

By symmetry x = -2 is also a solution.

Apr 9, 2018

x=2or x=-2

Explanation:

Here,

{sqrt(3)+sqrt(2)}^x + {sqrt(3)-sqrt(2)}^x =10

=>(sqrt3+sqrt2)^x+(((sqrt3-sqrt2) (sqrt3+sqrt2))/((sqrt3+sqrt2)))^x=10

=>(sqrt3+sqrt2)^x+((3-2)/((sqrt3+sqrt2)))^x=10

Taking, color(blue)((sqrt3+sqrt2)^x=m, we get

m+(1/m)=10

=>m^2+1=10m

=>m^2-10m=-1

=>m^2-10m+25=25-1=24

=>(m-5)^2=(2sqrt6)^2

=>m-5=+-2sqrt6

=>m=5+-2sqrt6

++=>m=(3+-2sqrt(3xx2)+2)

=>m=(sqrt3)^2+-2sqrt3sqrt2+(sqrt2)^2

=>m=(sqrt3+-sqrt2)^2

But ,we have taken color(blue)(m=(sqrt3+sqrt2)^x

So,

(sqrt3+sqrt2)^x=(sqrt3+sqrt2)^2

or(sqrt3+sqrt2)^x=(sqrt3-sqrt2)^2 =(1/(sqrt3+sqrt2))^2

i.e.(sqrt3+sqrt2)^x=(sqrt3+sqrt2)^2or(sqrt3+sqrt2)^x= (sqrt3+sqrt2)^-2

Comparing we get,

x=2or x=-2