Question

use a graphing tool to solve the equation below for x.
$-3^{(-x)} - 6 = -3^{x} + 10$
a. $x \approx 2.50$
b. $x \approx -6.00$
c. $x \approx -0.50$
d. $x \approx -2.50$

Explanation:

Step1: Rewrite the equation

First, rewrite \(-3^{(-x)}\) as \(-\frac{1}{3^{x}}\) (using the property \(a^{-n}=\frac{1}{a^{n}}\)). So the equation \(-3^{(-x)} - 6 = -3^{x}+10\) becomes \(-\frac{1}{3^{x}}-6=-3^{x} + 10\). Let \(y = 3^{x}\), then the equation is \(-\frac{1}{y}-6=-y + 10\). Multiply through by \(y\) (assuming \(y
eq0\)) to get \(- 1-6y=-y^{2}+10y\), which rearranges to \(y^{2}-16y - 1 = 0\). But maybe a better approach is to consider the two functions: \(f(x)=-3^{-x}-6\) and \(g(x)=-3^{x}+10\), and find their intersection point using a graphing tool.

Step2: Analyze the options (or use graphing logic)

We can test the options:

• For option A: \(x = 2.5\), \(3^{2.5}=3^{2+\frac{1}{2}}=9\sqrt{3}\approx15.588\). \(f(2.5)=-3^{-2.5}-6=-\frac{1}{15.588}-6\approx - 6.064\). \(g(2.5)=-15.588 + 10=-5.588\). Not equal.
• For option B: \(x=-6\), \(3^{-6}=\frac{1}{729}\approx0.00137\). \(f(-6)=-3^{6}-6=-729 - 6=-735\). \(g(-6)=-3^{-6}+10\approx - 0.00137+10 = 9.9986\). Not equal.
• For option C: \(x=-0.5\), \(3^{-0.5}=\frac{1}{\sqrt{3}}\approx0.577\). \(f(-0.5)=-3^{0.5}-6=-\sqrt{3}-6\approx - 7.732\). \(g(-0.5)=-3^{-0.5}+10\approx - 0.577+10 = 9.423\). Not equal.
• For option D: \(x = - 2.5\), \(3^{-2.5}=\frac{1}{3^{2.5}}=\frac{1}{9\sqrt{3}}\approx\frac{1}{15.588}\approx0.0641\). \(f(-2.5)=-3^{2.5}-6=-9\sqrt{3}-6\approx - 15.588-6=-21.588\)? Wait, no, wait: \(3^{-x}\) when \(x=-2.5\) is \(3^{2.5}\), so \(-3^{-x}=-3^{2.5}\). Wait, I made a mistake earlier. \(3^{-x}\) when \(x = - 2.5\) is \(3^{2.5}\), so \(f(x)=-3^{-x}-6=-3^{2.5}-6\approx - 15.588-6=-21.588\). \(g(x)=-3^{x}+10=-3^{-2.5}+10\approx-\frac{1}{15.588}+10\approx9.9359\). Wait, that's not matching. Wait, maybe my initial rewrite was wrong. Wait, the original equation: \(-3^{(-x)}-6=-3^{x}+10\). Let's re - express \(3^{-x}=\frac{1}{3^{x}}\), so the left side is \(-\frac{1}{3^{x}}-6\), right side is \(-3^{x}+10\). Let's let \(t = 3^{x}\), then the equation is \(-\frac{1}{t}-6=-t + 10\), multiply by \(t\): \(-1-6t=-t^{2}+10t\), \(t^{2}-16t - 1 = 0\). Using quadratic formula \(t=\frac{16\pm\sqrt{256 + 4}}{2}=\frac{16\pm\sqrt{260}}{2}=8\pm\sqrt{65}\). Since \(t = 3^{x}>0\), \(t = 8+\sqrt{65}\approx8 + 8.062=16.062\) (the other root is negative, discarded). Then \(3^{x}=16.062\), take log base 3: \(x=\log_{3}(16.062)=\frac{\ln(16.062)}{\ln(3)}\approx\frac{2.779}{1.0986}\approx2.53\). Wait, but the options have \(x\approx2.50\) (option A). Maybe my testing was wrong. Wait, maybe I messed up the function definitions. Let's define \(f(x)=-3^{-x}-6\) and \(g(x)=-3^{x}+10\). Let's use a graphing calculator approach: the intersection of \(y=-3^{-x}-6\) and \(y=-3^{x}+10\). When \(x = 2.5\), \(3^{2.5}\approx15.588\), so \(y_1=-3^{-2.5}-6\approx-\frac{1}{15.588}-6\approx - 6.064\), \(y_2=-15.588 + 10=-5.588\). Close? Wait, maybe the approximate value is around 2.5. Alternatively, the correct answer is A as the closest approximation.

Answer:

A. \(x\approx2.50\)