Question

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

Answer 1

`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.

Answer 2

`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`