Solving $4^{x+\sqrt{x^2-2}}-5\cdot 2^{\sqrt{x^2-2}-1}=6$ for real $x$

Question

I was given the following equation: $$4^{x+\sqrt{x^2-2}}-5\cdot 2^{\sqrt{x^2-2}-1}=6; x\in \mathbb{R}$$

Here is how I tried to solve it:

$$2^{2x+2\sqrt{x^2-2}+1}-5\cdot 2^{\sqrt{x^2-2}}=12$$

$$2^{\sqrt{x^2-2}}(2^{2x+1+\sqrt{x^2-2}}-5)=2^2\cdot 3$$

Where $$2^{\sqrt{x^2-2}}=2^2 \land 2^{2x+1+\sqrt{x^2-2}}-5=3$$ $$\Updownarrow$$ $$\sqrt{x^2-2}=2 \land 2x+1+\sqrt{x^2-2}=3.$$

As well as we must note that $$x\in \left( -\infty ; -\sqrt2 \right]\cup \left[ \sqrt{2};\infty\right)$$

However

$$2x+1+\sqrt{x^2-2}=3 \Leftrightarrow -3x^2+8x-6=0$$ $$x_{1,2}=\frac{8\pm\sqrt{-8}}{6}$$

Thus, it's not satisfying given domain. So I tried to use substitution

Let $\sqrt{x^2-2}=t, t\geq 0,$ then $x^2=t^2+2 \Leftrightarrow x=\pm \sqrt{t^2+2}$

If $$x=\sqrt{t^2+2}$$ then $$4^{x+\sqrt{x^2-2}}-5\cdot 2^{\sqrt{x^2-2}-1}=6$$ $$\Updownarrow$$ $$2^{2\sqrt{t^2+2}+2t}-5\cdot 2^{t-1}=6$$ $$\Updownarrow$$ $$2^{2\sqrt{t^2+2}+2\sqrt{t^2}}-5\cdot 2^{\sqrt{t^2}-1}=6$$

From here if I continued using substitution to make it into a quadratic equation, I would reach a result, however, I doubt this is the right way as I only had 10 minutes to solve this in class and I'm only up to high school. Can someone provide me with any solution?

Answer 1

Comment: As you modified we have:

$2^{2x+1+2\sqrt{x^2-2}}-5\cdot 2^{\sqrt {x^2-2}}-12=0$

$2^{2x+1+2\sqrt{x^2-2}}=2^{2x+1}\cdot 2^{2\sqrt {x^2-2}}=2^{2x+1}\cdot (2^{\sqrt{x^2-2}})^2$

$2^{2x+1}\cdot (2^{\sqrt{x^2-2}})^2-5\cdot 2^{\sqrt{x^2-2}}-12=0$

Let $2^{\sqrt{x^2-2}}=y$ , we can write:

$(2^{2x+1})\cdot y^2-5y-12=0$

$y=\frac {5\pm\sqrt{25+48\cdot 2^{2x+1}}}{2\times 2^{2x+1}}$

Now you have to search whether following system of equations is consistent or not:

$\begin {cases}\Delta=25+48\cdot 2^{2x+1}=k^2\\y=2^{\sqrt{x^2-2}}\end {cases}$

For solutions in $ R$ we take only positive solutions for y:

$y=\frac {5+\sqrt{25+48\cdot 2^{2x+1}}}{2\times 2^{2x+1}}$

Update: we can see that:

$2^{2x+1}=11\rightarrow 2x+1\approx 3.5\rightarrow x=1.25$

$y=\frac{5+17}{2\times 11}=1=2^{\sqrt{x^2-2}}\Rightarrow X^2-2=0$

which gives $x=\sqrt 2\approx 1.4...$

1- $\sqrt 2$ gives $LHS=5.6$

2-$x=2$, gives $LHS=39.3$

3- In fact we must have $\sqrt 2<x< 2$, $x=1.6$ works:

$LHS\approx 12$

No need a complicated analytic solution; these three steps can be done using a calculator in less that ten minutes. What is more important is to find the domain.