Example Question 5

Example corresponding to question 5 in college algebra 2.

Example 1


If $(4^{-x^2})(16^{-2x})=\dfrac{1}{2^{-8}}$, then $x=?$

  1. $-1$


  2. $2$


  3. $-2$


  4. $4$


  5. $-4$

Solution


  1. 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

    $((2^2)^{-x^2})((2^4)^{-2x})=\dfrac{1}{2^{-8}}$

  2. We now use the exponential rule $(x^m)^n = x^{m \cdot n}$ and definition of negative exponents to simplify the equation as follows

    $(2^{-2x^2})(2^{-8x})=2^8$

  3. Group the terms on the left using the rule $x^m \cdot x^n = x^{m+n}$

    $2^{-2x^2-8x}=2^8$

  4. Since the exponential function is a one to one function, the above gives

    $-2x^2-8x = 8$

  5. Divide all terms of the above equation by $2$ and rewrite in standard form.

    $x^2+4x + 4=0$

  6. Factor and solve

    $(x+2)^2=0$

    Solution: $x = -2$

    Answer C