Question

log₇(x) - 1 - log₇(x)² + log₇(x²) divided by log₇(x)² + log₇(x²) ≤ 0

Explanation:

Step 1: Simplify the logarithm term \(\log_7(x^2)\)

Using the logarithm power rule \(\log_a(b^c)=c\log_a(b)\), we have \(\log_7(x^2) = 2\log_7(x)\). Let \(t=\log_7(x)\), then the inequality becomes:

$$ \frac{t - 1 - t^2+2t}{t^2 + 2t}\leq0 $$

Simplify the numerator: \(t - 1 - t^2+2t=-t^2 + 3t - 1\)
So the inequality is \(\frac{-t^2 + 3t - 1}{t^2 + 2t}\leq0\), multiply numerator and denominator by - 1 (note that this will reverse the inequality sign):

$$ \frac{t^2 - 3t + 1}{t^2 + 2t}\geq0 $$

Factor the denominator: \(t^2 + 2t=t(t + 2)\)
The roots of the numerator \(t^2-3t + 1 = 0\) are given by the quadratic formula \(t=\frac{3\pm\sqrt{9 - 4}}{2}=\frac{3\pm\sqrt{5}}{2}\approx\frac{3\pm2.24}{2}\), so \(t_1=\frac{3 + \sqrt{5}}{2}\approx2.62\) and \(t_2=\frac{3-\sqrt{5}}{2}\approx0.38\)
The roots of the denominator are \(t = 0\) and \(t=-2\)

Step 2: Analyze the sign of the function \(y = \frac{t^2 - 3t + 1}{t(t + 2)}\)

We consider the intervals determined by the critical points \(t=-2\), \(t = 0\), \(t=\frac{3-\sqrt{5}}{2}\), \(t=\frac{3+\sqrt{5}}{2}\)

• For \(t<-2\), let \(t=-3\), numerator \(=9 + 9+1 = 19>0\), denominator \(=(-3)\times(-1)=3>0\), so \(y>0\)
• For \(-20\), denominator \(=(-1)\times1=-1<0\), so \(y<0\)
• For \(00\), denominator \(=0.2\times2.2 = 0.44>0\), so \(y>0\)
• For \(\frac{3 - \sqrt{5}}{2}0\), so \(y<0\)
• For \(t>\frac{3+\sqrt{5}}{2}\), let \(t = 3\), numerator \(=9-9 + 1 = 1>0\), denominator \(=3\times5 = 15>0\), so \(y>0\)

We want \(\frac{t^2 - 3t + 1}{t(t + 2)}\geq0\), so the solution for \(t\) is \(t<-2\) or \(0

Step 3: Convert back to \(x\) (since \(t=\log_7(x)\))

• Case 1: \(t<-2\), i.e., \(\log_7(x)<-2=\log_7(7^{-2})=\log_7(\frac{1}{49})\)

Since the function \(y = \log_7(x)\) is increasing, \(x<\frac{1}{49}\), but we also need \(x>0\) (domain of logarithm), so \(0

• Case 2: \(0

Since \(y = \log_7(x)\) is increasing, \(7^0

• Case 3: \(t\geq\frac{3+\sqrt{5}}{2}\), i.e., \(\log_7(x)\geq\frac{3+\sqrt{5}}{2}\)

Since \(y=\log_7(x)\) is increasing, \(x\geq7^{\frac{3+\sqrt{5}}{2}}\)

Answer:

The solution of the inequality is \(0