Solving the logarithmic inequality by using manipulation

Question

I am trying to solve this problem but stuck $$\log_x\frac{4x+5}{6-5x}<-1$$

Answer 1

We have that for $x>0$, $x\not=1$, and $A>0$, $$\log_x(A)=\frac{\ln (A)}{\ln(x)}$$ where $\ln$ is the natural logarithm.

In our case, for $x>0$, $x\not=1$, $$A=\frac{4x+5}{6-5x}>0\Leftrightarrow x\in (0,1)\cup (1,6/5).$$

Now we have two cases.

1) If $1<x<6/5$, then $\ln(x)>0$ and the inequality is equivalent to $$\ln \frac{4x+5}{6-5x}<-\ln(x)=\ln(1/x) \Leftrightarrow \frac{4x+5}{6-5x}<\frac{1}{x} \Leftrightarrow 4x^2+5x < 6−5x \Leftrightarrow 4x^2+10x < 6 $$ which is impossible for $x>1$.

2) If $0<x<1$, then $\ln(x)<0$ and the inequality is equivalent to $$\ln\frac{4x+5}{6-5x}>-\ln(x)=\ln(1/x) \Leftrightarrow \frac{4x+5}{6-5x}>\frac{1}{x}\Leftrightarrow 4x^2+10x >6\Leftrightarrow x\in (1/2,1).$$

We conclude that the inequality is satisfied iff $x\in (1/2,1)$.

Answer 2

Can you take the log in a negative base?

While we are on the subject of undefined solutions, can we take the log of a negative number. Can, $\frac {4x+5}{6−5x} < 0$

And what would be the implications of $x = 1$? That would also be undefined

The domain of $\log_x \frac {4x+5}{6−5x}$ is $x\in(0,1)\cup(1,\frac 65)$

$\log_x \frac {4x+5}{6−5x}<(−1)$

Take the inverse log of both sides. What is going to happen to that inequality? Does it stay the same, or does if flip?

If $x>1, x^a < x^b$ when $a<b$ when $x>1, x^a>x^b$ when $a<b$. We must consider the two cases separately.

case $x>1:$ $\frac {4x+5}{6−5x}<x^{−1}\\ \frac {4x+5}{6−5x}<\frac {1}{x}\\ 4x^2+5x < 6−5x\\ (4x - 2)(x+3)<0\\ x\in(-3,\frac12)$

But that contradicts that $x>1$

Note: when cross multiplying watch for the possibilities of flipping the inequalities when $x<0.$ Not the case here, so not a risk.

Case $0<x<1:$ $\frac {4x+5}{6−5x}>x^{−1}\\ \frac {4x+5}{6−5x}>\frac {1}{x}\\ 4x^2+5x > 6−5x\\ (4x - 2)(x+3)>0\\ x>\frac 12$

$x\in (\frac12,1)$

Answer 3

Hint:

Note that $$\log_x\frac{4x+5}{6-5x}$$ is a real number iff $0<x<\frac{6}{5}$.

We know that $y=\log_a x$ is a monotoning increasing function for $a>1$ and motonotic decreasing if $0<a<1$.

So, from the definition: $\log_x A=-1 \iff x^{-1}=A $, we have two cases for the given inequality:

$$ \begin {cases} 1<x<\frac{6}{5}\\ \frac{4x+5}{6-5x}<x^{-1} \end{cases} $$

$$ \begin {cases} 0<x<1\\ \frac{4x+5}{6-5x}>x^{-1} \end{cases} $$

Answer 4

Let $f(x):=\frac{4x+5}{6-5x}.$

Firstly, let us work on domain restrictions:

• 1) $f(x)$ must be $>0$, (because the domain of any logarithm function is $(0,\infty)$; thus $-5/4<x<6/5.$

• 2) $x>0$ (because any basis of logarithms is positive).

Therefore, values of $x$ are restricted to open interval $I:=(0,6/5)$.

Taking the definition of a logarithm in base $b$:

$$\tag{1}\dfrac{\ln(f(x))}{\ln(x)}<-1 \Leftrightarrow \begin{cases}(a) &if& 0<x<1&\ln(f(x))>-\ln(x)\Leftrightarrow f(x)>\frac{1}{x}\\(b) &if& 1<x<6/5&\ln(f(x))<-\ln(x)\Leftrightarrow f(x)<\frac{1}{x}\end{cases}$$

(we have used the increasing property of function $\ln$).

Can you conclude from there?