Two exponential inequalities

Statement

Solve the following exponential inequalities:

• (a) $3-\dfrac{1}{2}\left(\dfrac{2}{3}\right)^x\ge 1$
• (b) $\dfrac{2}{3}\left(\dfrac{1}{2}\right)^{1-x^2}-1\ge 0$

Solution

(a) Isolate the power: $-\tfrac12\left(\tfrac23\right)^x\ge -2\Rightarrow\left(\tfrac23\right)^x\le 4$. The base is $<1$, so the inequality reverses: $x\ge\log_{2/3}4\approx -3.42$.

(b) $\left(\tfrac12\right)^{1-x^2}\ge\tfrac32$. The base is $<1$: taking the logarithm reverses the inequality, $1-x^2\le\log_{1/2}\tfrac32\approx -0.585$, hence $x^2\ge 1.585\Rightarrow |x|\ge 1.26$. That is $x\le -1.26\ \vee\ x\ge 1.26$.

$\boxed{\text{(a) }x\ge\log_{2/3}4\approx -3.42;\quad \text{(b) }x\le -1.26\ \vee\ x\ge 1.26}$