Question

16^х - 18 * 4^х+ 32 = 0 РЕШИТЕЕЕЕЕЕЕЕЕЕЕЕЕЕЕЕЕЕЕЕЕЕЕЕЕЕЕЕЕЕЕЕЕ

Answer 1

16^x-18*4^x+32=0
4^2x-18*4^x+32=0
Пусть 4^x=t
t^2-18*t+32=0
t1=16 t2=2 (по теореме Виета)
4^x=16 4^x=2
x=2 x=0,5
Ответ:0,5;2.

Answer 2

To solve the equation $16^x - 18 \cdot 4^x + 32 = 0$, we can make a substitution to simplify the equation. Let $y = 4^x$, then the equation becomes:

$y^2 - 18y + 32 = 0$

Now, we can factor this quadratic equation:

$(y - 16)(y - 2) = 0$

This gives two possible solutions for $y$:

1. $y - 16 = 0$ $\Rightarrow$ $y = 16$
2. $y - 2 = 0$ $\Rightarrow$ $y = 2$

Now, remember that $y = 4^x$. So, we have:

1. $4^x = 16$ $\Rightarrow$ $2^{2x} = 2^4$ $\Rightarrow$ $2x = 4$ $\Rightarrow$ $x = 2$
2. $4^x = 2$ $\Rightarrow$ $2^{2x} = 2^1$ $\Rightarrow$ $2x = 1$ $\Rightarrow$ $x = 0.5$

So, the solutions to the equation $16^x - 18 \cdot 4^x + 32 = 0$ are $x = 2$ and $x = 0.5$.