Question

The solution set of ln(5-7x)<=1 is given by

Answer 1

Given inequality is,

$\longrightarrow\ln(5-7x)\leq1$

Taking antilog,

$\longrightarrow e^{\ln(5-7x)}\leq e^1$

$\longrightarrow 5-7x\leq e$

Multiplying both sides by -1, (Note the sign change)

$\longrightarrow 7x-5\geq -e$

Adding 5,

$\longrightarrow 7x\geq5-e$

Dividing by 7, we get,

$\longrightarrow x\geq\dfrac{5-e}{7}$

$\Longrightarrow x\in\left[\dfrac{5-e}{7},\ \infty\right)\quad\quad\dots(1)$

But we know $\ln(x)$ is only defined for $x>0.$

So for $\ln(5-7x)$ being defined,

$\longrightarrow 5-7x>0$

Multiplying both sides by -1, (Note the sign change)

$\longrightarrow 7x-5<0$

Adding 5,

$\longrightarrow 7x<5$

Dividing by 7, we get,

$\longrightarrow x<\dfrac{5}{7}$

$\Longrightarrow x\in\left(-\infty,\ \dfrac{5}{7}\right)\quad\quad\dots(2)$

Combining (1) and (2),

$\longrightarrow x\in\left(-\infty,\ \dfrac{5}{7}\right)\cap\left[\dfrac{5-e}{7},\ \infty\right)$

$\longrightarrow\underline{\underline{x\in\left[\dfrac{5-e}{7},\ \dfrac{5}{7}\right)}}$

This is the solution set of the inequality.