Where am I wrong?

Question

I need to solve the inequality $5^x-10 ≥ \frac{225}{5x-10}$. Here's my (incorrect) solution: $$ 5^x ≥ \frac{225+50x-100}{5x-10} $$ $$5^x ≥ \frac{50x+125}{5x-10}$$ $$5^x ≥ \frac{5(10x+25)}{5(x+2)}$$ $$5^x ≥ \frac{25(2x+5)}{5(x+2)}$$ $$5^x ≥ \frac{5(2x+5)}{x+2}$$ $$5^{x-1} ≥ 5^{\log_5\frac{10x+25}{x+2}}$$ $$x-1 ≥ \log_5\frac{10x+25}{x+2}$$ $$\log_5\frac{2x+5}{x+2}≥\log_5\frac{10x+25}{x+2}$$ $$\frac{2x+5-10x-25}{x+2}≥0$$ $$\frac{-20-8x}{x+2}≥0$$ If my solution were correct, then the answer should be the interval [-2.5; -2), but at $x = -2.5$ $5^x-10 ≠ \frac{225}{5x-10}$, which implies that my solution is incorrect. Where are the errors in my solution?

Answer 1

As noticed in the comments you have made a mistake at the third step, note also that when we take logarithm we need to satisfy some condition.

Notably, we can proceed as follows

$$5^x ≥ \frac{225+50x-100}{5x-10} \iff 5^x ≥ \frac{10x+25}{x-2}$$

Now since LHS is always positive the inequality is true when

$$\frac{10x+25}{x-2}\le 0 \implies x\in\left[-\frac52, 2\right)$$

otherwise when $x \in \left( -\infty,-\frac52 \right)\cup \left( 2,\infty \right)$ we can observe that $5^x$ is a strictly increasing function while the RHS is a strictly decreasing function on the interval, therefore it suffices to consider the equation

$$5^x = \frac{10x+25}{x-2}$$

then show that two real solutions exist (by IVT) and find them by numerical methods.

Answer 2

Too long for a comment.

As @user wrote, to finish the problem, you need to solve for $x$ the transcendental equation

$$5^x = \frac{10x+25}{x-2}$$ which happens to have an explicit solution in terms of the generalized Lambert function

Rewrite is as $$e^{-x\log(5)}=\frac 1{10}\,\frac {x-2}{x+\frac 52 }$$ and have a look at equation $(4)$ in the linked paper.

This is nice from a formal point of view but more than likely not very practical for you at the time being.

Cheating a little, plotting, you will notice that the solutions are close to $x=\pm \frac 52$.

So, instead of using a numerical method, let $x=\pm\frac 52+\epsilon$ and expand as a Taylor series limited to the first term around $\epsilon=0$

For the negative root $$5^{\epsilon -\frac{5}{2}}+\frac{20 \epsilon }{9-2 \epsilon}=\frac{1}{25 \sqrt{5}}+ \left(\frac{20}{9}+\frac{\log(5)}{25 \sqrt{5}}\right)\epsilon+O\left(\epsilon^2\right)$$ gives $$\epsilon \sim -\frac{9 \sqrt{5}}{2500+9 \sqrt{5} \log(5)} =\color{red}{-0.0079}47\quad \implies \quad x_- \sim -2.5079$$

For the positive root $$5^{\epsilon +\frac{5}{2}}-\frac{20 (\epsilon +5)}{2 \epsilon +1}=-25 \left(4-\sqrt{5}\right)+5\left(36+5 \sqrt{5} \log(5)\right) \epsilon +O\left(\epsilon ^2\right)$$ gives $$\epsilon \sim \frac{5 \left(4-\sqrt{5}\right)}{36+5\sqrt{5} \log (5)}=0.163345$$ while the solution is $0.186861$.