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

Apr 9, 2018

$\pm 2$

#### Explanation:

Calling

$a = \sqrt{3} + \sqrt{2}$
$b = \sqrt{3} - \sqrt{2}$

we have the equivalent problem

$\left\{\begin{matrix}{a}^{x} + {b}^{x} = 10 \\ a b = 1\end{matrix}\right.$

or

${a}^{x} + {a}^{-} x = 10$ or

${e}^{\lambda x} + {e}^{- \lambda x} = 10$ or

$2 \cosh \left(\lambda x\right) = 10$ or

$x = \frac{1}{\lambda} \text{arccosh} \left(5\right)$

Here $\lambda = \ln a = \ln \left(\sqrt{3} + \sqrt{2}\right)$ so finally

$x = \frac{\text{arccosh} \left(5\right)}{\ln} \left(\sqrt{3} + \sqrt{2}\right) = 2$

By symmetry $x = - 2$ is also a solution.

Apr 9, 2018

$x = 2 \mathmr{and} x = - 2$

#### Explanation:

Here,

${\left\{\sqrt{3} + \sqrt{2}\right\}}^{x} + {\left\{\sqrt{3} - \sqrt{2}\right\}}^{x} = 10$

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

$\implies {\left(\sqrt{3} + \sqrt{2}\right)}^{x} + {\left(\frac{3 - 2}{\left(\sqrt{3} + \sqrt{2}\right)}\right)}^{x} = 10$

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

$m + \left(\frac{1}{m}\right) = 10$

$\implies {m}^{2} + 1 = 10 m$

$\implies {m}^{2} - 10 m = - 1$

$\implies {m}^{2} - 10 m + 25 = 25 - 1 = 24$

$\implies {\left(m - 5\right)}^{2} = {\left(2 \sqrt{6}\right)}^{2}$

$\implies m - 5 = \pm 2 \sqrt{6}$

$\implies m = 5 \pm 2 \sqrt{6}$

++$\implies m = \left(3 \pm 2 \sqrt{3 \times 2} + 2\right)$

$\implies m = {\left(\sqrt{3}\right)}^{2} \pm 2 \sqrt{3} \sqrt{2} + {\left(\sqrt{2}\right)}^{2}$

$\implies m = {\left(\sqrt{3} \pm \sqrt{2}\right)}^{2}$

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

So,

${\left(\sqrt{3} + \sqrt{2}\right)}^{x} = {\left(\sqrt{3} + \sqrt{2}\right)}^{2}$

$\mathmr{and} {\left(\sqrt{3} + \sqrt{2}\right)}^{x} = {\left(\sqrt{3} - \sqrt{2}\right)}^{2} = {\left(\frac{1}{\sqrt{3} + \sqrt{2}}\right)}^{2}$

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

Comparing we get,

$x = 2 \mathmr{and} x = - 2$