Question

solve the inequality \\(\frac{\log_{7}(49x) - 3}{\log_{7}^{2}x + \log_{7}x^{2}} \leq 1\\).

Explanation:

Step1: Simplify the numerator and denominator

First, simplify \(\log_{7}(49x)\). We know that \(\log_{7}(49x)=\log_{7}(49)+\log_{7}(x) = 2+\log_{7}(x)\) (using the property \(\log_a(MN)=\log_aM+\log_aN\) and \(\log_7 49 = 2\) since \(7^2 = 49\)).

For the denominator, \(\log_{7}^{2}x+\log_{7}x^{2}=\log_{7}^{2}x + 2\log_{7}x\) (using the property \(\log_aM^n=n\log_aM\)). Let \(t = \log_{7}x\), then the inequality becomes:

\[ \frac{(2 + t)-3}{t^{2}+2t}\leq1 \]

Simplify the numerator: \(2 + t-3=t - 1\). So the inequality is \(\frac{t - 1}{t^{2}+2t}\leq1\)

Step2: Rearrange the inequality

Subtract 1 from both sides:

\[ \frac{t - 1}{t^{2}+2t}-1\leq0 \]

Find a common denominator \(t^{2}+2t=t(t + 2)\) (note that \(t eq0\) and \(t eq - 2\) to avoid division by zero):

\[ \frac{t - 1-(t^{2}+2t)}{t(t + 2)}\leq0 \]

Simplify the numerator: \(t - 1-t^{2}-2t=-t^{2}-t - 1=-(t^{2}+t + 1)\)

The discriminant of \(t^{2}+t + 1\) is \(\Delta=1^{2}-4\times1\times1=1 - 4=- 3<0\), and the coefficient of \(t^{2}\) is positive, so \(t^{2}+t + 1>0\) for all real \(t\). Then \(-(t^{2}+t + 1)<0\) for all real \(t\)

So the inequality \(\frac{-(t^{2}+t + 1)}{t(t + 2)}\leq0\) is equivalent to \(\frac{-(\text{negative})}{t(t + 2)}\leq0\) (since \(t^{2}+t + 1>0\)) or \(\frac{\text{negative}}{t(t + 2)}\leq0\), which is equivalent to \(\frac{1}{t(t + 2)}\geq0\) (multiplying both sides by - 1 reverses the inequality)

Step3: Solve the inequality \(\frac{1}{t(t + 2)}\geq0\)

The function \(y = \frac{1}{t(t + 2)}\) has vertical asymptotes at \(t = 0\) and \(t=-2\). We consider the intervals:

- For \(t<-2\): \(t<0\) and \(t + 2<0\), so \(t(t + 2)>0\), then \(\frac{1}{t(t + 2)}>0\)
- For \(-2<t<0\): \(t<0\) and \(t + 2>0\), so \(t(t + 2)<0\), then \(\frac{1}{t(t + 2)}<0\)
- For \(t>0\): \(t>0\) and \(t + 2>0\), so \(t(t + 2)>0\), then \(\frac{1}{t(t + 2)}>0\)

We exclude \(t = 0\) and \(t=-2\) because the original function is undefined there. So the solution for \(t\) is \(t<-2\) or \(t>0\)

Step4: Substitute back \(t=\log_{7}x\)

Case 1: \(\log_{7}x<-2\)

Using the property of logarithms, \(\log_{7}x<-2=\log_{7}7^{-2}=\log_{7}\frac{1}{49}\)

Since the function \(y = \log_{7}u\) is an increasing function (because the base \(7>1\)), we have \(x<7^{-2}=\frac{1}{49}\). But we also need to consider the domain of the original logarithm function, \(x>0\). So for this case, \(0<x<\frac{1}{49}\)

Case 2: \(\log_{7}x>0=\log_{7}1\)

Since \(y = \log_{7}u\) is increasing, \(x>1\)

Step5: Check the original inequality for domain restrictions

We also need to check the original denominator \(\log_{7}^{2}x+\log_{7}x^{2} eq0\)

\(\log_{7}^{2}x+2\log_{7}x=\log_{7}x(\log_{7}x + 2) eq0\)

Which means \(\log_{7}x eq0\) (i.e., \(x eq1\)) and \(\log_{7}x eq - 2\) (i.e., \(x eq\frac{1}{49}\))

From Case 1: \(0<x<\frac{1}{49}\) (since \(x eq\frac{1}{49}\) and \(x>0\))

From Case 2: \(x>1\) (since \(x eq1\))

We also need to check the original numerator \(\log_{7}(49x)-3\) and the denominator \(\log_{7}^{2}x+\log_{7}x^{2}\) for the original inequality.

Let's verify with a test point in \(0<x<\frac{1}{49}\), say \(x=\frac{1}{343}\)

\(\log_{7}(49\times\frac{1}{343})=\log_{7}(\frac{49}{343})=\log_{7}(\frac{1}{7})=- 1\)

Numerator: \(-1-3=-4\)

Denominator: \(\log_{7}^{2}(\frac{1}{343})+\log_{7}(\frac{1}{343})^{2}\)

\(\log_{7}(\frac{1}{343})=\log_{7}(7^{-3})=-3\), so \(\log_{7}^{2}(\frac{1}{343})=9\) and \(\log_{7}(\frac{1}{343})^{2}=\log_{7}(7^{-6})=-6\), denominator \(=9-6 = 3\)

\(\frac{-4}{3}\leq1\), which is true.

For a test point in \(x > 1\), say \(x = 7\)

\(\log_{7}(49\times7)=\log_{7}(343)=\log_{7}(7^{3})=3\)

Numerator: \(3 - 3=0\)

Denominator: \(\log_{7}^{2}(7)+\log_{7}(7^{2})=1 + 2=3\)

\(\frac{0}{3}=0\leq1\), which is true.

For \(x = 1\), the original denominator: \(\log_{7}^{2}(1)+\log_{7}(1)=0 + 0=0\), so \(x = 1\) is not in the domain.

For \(x=\frac{1}{49}\), \(\log_{7}(\frac{1}{49})=-2\), denominator: \(\log_{7}^{2}(\frac{1}{49})+\log_{7}(\frac{1}{49})^{2}=4+\log_{7}(7^{-4})=4-4 = 0\), so \(x=\frac{1}{49}\) is not in the domain.

Answer:

The solution of the inequality is \(0 1\)