# Thread: I can't find the root(s) of this equation (indices)

1. ## I can't find the root(s) of this equation (indices)

Here are my workings solve the equation $2^{2x-1} =16 +2^{x+1}$ :

2. Dear PythagorasNeophyte,

Your quadratic equation is incorrect. It should be, $y^2-4y-32=0~where~y=2^x$.

3. Your error was in going from

$2^x2^x2^{-1}=2^4+(2)2^x$ to $2^x2^x=2^5+(4)2^x$

You inadvertently switched the addition and multiplication symbols on the right-hand side.

You could also try

$2^{2x-1}-2^{x+1}=2^4$

$2^{2x}-2^{x+2}=2^5$

$2^x\left[2^x-4\right]=32$

The only two positive numbers that differ by 4 and whose product is 32 are 8 and 4 so by inspection $2^x=8$

4. Hello, PythagorasNeophyte!

Sheesh . . . you're at this level of algebra
. . and you're still using $\times$ for multiplication?

$2^{2x-1} \:=\:16 +2^{x+1}$

We have: . $2^{2x-1} - 2^{x+1} - 16 \:=\:0$

. . . . . $2^{2x}\cdot2^{-1} - 2^x\cdot 2 - 16 \:=\:0$

. . . . . . . $\frac{1}{2}\!\cdot\!2^{2x} - 2\cdot2^x - 16 \:=\:0$

Multiply by 2: . $2^{2x} - 4\cdot2^x - 32 \:=\:0$

Let $y = 2^x\!:\;\;y^2 - 4y - 32 \:=\:0\quad\Rightarrow\quad (y + 4)(y-8) \:=\:0 \quad\Rightarrow\quad y \:=\:-4,\:8$

Back-substitute: . $\begin{Bmatrix} 2^x \:=\:\text{-}4 & \Rightarrow & \text{no real roots} \\
2^x \:=\:8 & \Rightarrow & \boxed{x \:=\:3} \end{Bmatrix}$

5. If you want a fancy solution

$2^{2x-1}=2^4+2^{x+1}$

divide both sides by 2 to obtain $2^x$ on the right

$2^{2x-2}=2^3+2^x$

$2^{2x-2}-2^3=2^x$

Divide both sides by $2^x$

$2^{x-2}-2^{3-x}=1$

The only powers of 2 whose index moduli and values differ by 1 are $2^0=1$ and $2^1=2$

$\Rightarrow\ x=3$