Fractions and Decimals

Discrete Mathematics

Differential Equations

Solved on Jan 19, 2024

Solve the equation $2^{(x-1)(x+2)(x-7)(x+8)}=1$ .

STEP 1

Assumptions
1. We are given the equation $2^{(x-1)(x+2)(x-7)(x+8)}=1$ .
2. We need to solve for the variable $x$ .

STEP 2

Recognize that any number raised to the power of 0 is equal to 1. Therefore, we can set the exponent of 2 to 0 to solve for $x$ .
$2^{(x-1)(x+2)(x-7)(x+8)}=1 \Rightarrow (x-1)(x+2)(x-7)(x+8) = 0$

STEP 3

We have a product of factors equal to zero. According to the zero-product property, if the product of several factors is zero, at least one of the factors must be zero.

STEP 4

Set each factor equal to zero to find the values of $x$ .
$x-1=0, \quad x+2=0, \quad x-7=0, \quad x+8=0$

STEP 5

Solve each equation for $x$ .
1. $x-1=0 \Rightarrow x=1$
2. $x+2=0 \Rightarrow x=-2$
3. $x-7=0 \Rightarrow x=7$
4. $x+8=0 \Rightarrow x=-8$

STEP 6

List all the solutions for $x$ .
The solutions are $x=1$ , $x=-2$ , $x=7$ , and $x=-8$ .

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