solve an exponential inequality $2a^x-a^{x+1} \le b$

Question

Given two scalars $0<a<1$ and $0<b<1$, how to solve the unknown $x$ in the following equation:

$2a^x-a^{x+1} \le b$

Answer 1

Let: $$2a^x-a^{x+1}=b$$ Then, \begin{align} (2-a)a^x&=b\\ a^x&=\frac{b}{2-a}\\ x &= \log_a\left(\frac{b}{2-a}\right)\\ x &= \log_a(b)-log_a(2-a))\\ \end{align} Fil in $x=0\leq \log_a(b)-log_a(2-a))$ to check the original inequality. We find $2a^0-a^1=2-a \geq b$, which does not satisfy. So $x\geq\log_a(b)-log_a(2-a))$ satisfies the original inequality.

Answer 2

You should notice that you can write the variable part as just $a^x$: $$2a^x-a^{x+1} \leq b$$ $$2a^x-\underbrace{a\cdot a^x}_{a^{x+1}} \leq b$$ $$(2-a)\cdot a^x \leq b$$ Since $0<a<1$, you know $2-1<2-a<2-0$, that is, $1<2-a<2$. In particular, $2-a$ is a positive number, so you can divide both sides of the inequality by $2-a$ without worrying about reversing the direction of the inequality: $$a^x \leq\frac{b}{2-a}$$ Since $\ln$ is an increasing function, it will also preserve the inequality: $$\ln a^x \leq\ln\frac{b}{2-a}$$ $$x\ln a \leq \ln\frac{b}{2-a}$$ Finally, since $0<a<1$, you know that $\ln a < 0$, so dividing both sides of the inequality by $\ln a$ will reverse the direction of the inequality: $$x\geq \frac{\ln\frac{b}{2-a}}{\ln a}=\log_a\frac{b}{2-a}$$ So the solution set is $$\boxed{x\geq \log_a\frac{b}{2-a}}$$