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Example 1
If $(4^{-x^2})(16^{-2x})=\dfrac{1}{2^{-8}}$, then $x=?$
- $-1$
- $2$
- $-2$
- $4$
- $-4$
Solution
- We first need to rewrite all exponential terms using the same base by substituting $4$ by $2^2$ and $16$ by $2^4$ in the given equation
$((2^2)^{-x^2})((2^4)^{-2x})=\dfrac{1}{2^{-8}}$
- We now use the exponential rule $(x^m)^n = x^{m \cdot n}$ and definition of negative exponents to simplify the equation as follows
$(2^{-2x^2})(2^{-8x})=2^8$
- Group the terms on the left using the rule $x^m \cdot x^n = x^{m+n}$
$2^{-2x^2-8x}=2^8$
- Since the exponential function is a one to one function, the above gives
$-2x^2-8x = 8$
- Divide all terms of the above equation by $2$ and rewrite in standard form.
$x^2+4x + 4=0$
- Factor and solve
$(x+2)^2=0$
Solution: $x = -2$
Answer C
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