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

2 Answers
Apr 9, 2018

Answer:

#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

Answer:

#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#