exponential equation to solve ($2^{x+1}\ + 3) (2^{x-1} -5) = -19 $

Question

($2^{x+1}\ + 3) (2^{x-1} -5) = -19 $

I multiplied and got:

$2^{2x}\ - 5 2^{x+1} + 3 2^{x-2} - 15 = 19 $

And then, I subtracted 19 from both sides to get:

$2^{2x}\ - 5 2^{x+1} + 3 2^{x-2} + 4 = 0 $

I got stuck here. Thank you for your help.

Answer 1

Hint: Make the substitution $y = 2^{x - 1}$. What does $2^{x + 1}$ become?

---

Full solution: We get $$ (4y + 3)(y - 5) = -19. $$ Expanding and rearranging gives $$ 4 y^2 - 17 y + 4 = 0 \implies y^2 - \frac{17}{4}y + 1 = 0 $$ yielding $$ y_{1} = \frac{1}{4} \qquad \text{and} \qquad y_2 = 4. $$

Undoing the substitution we now have for $i \in \{1,2\}$ (and $y > 0$) $$ \tag{$\star$} y_i = 2^{x_i - 1} \iff x_i = \log_2(y_i) + 1 $$ and therefore we get \begin{gather} x_1 = \log_2\left(\frac{1}{4}\right) + 1 = \log_2(2^{-2}) + 1 = -1 \\ x_2 = \log_2(4) + 1 = \log_2(2^2) + 1 = 3. \end{gather}

Answer 2

$2^{2x}-5\times2^{x+1}+3\times2^{x-1}-15=-19\implies2^{2x}-10\times2^x+\dfrac32\times2^x+4=0$

$\implies y^2-\dfrac{17}2y+4=0$, where $y=2^x$ as suggested by Lord Shark the Unknown.

Can you take it from here?