**Logarithmic inequalities** are inequalities in which one or both sides contain a logarithm. While solving logarithmic inequalities, we must keep in mind these facts:

**1)** If $a>1$ and $x<y$, then $log_{a}x<log_{a}y$.

If $a>1$ and $log_{a}x<log_{a}y$, then $x<y$.

Precisely, the logarithmic function $f(x)=log_{a}x$ is **monotonically increasing** for **$a>1$**.

**2)** If $0<a<1$ and $x<y$, then $log_{a}x>log_{a}y$.

If $0<a<1$ and $log_{a}x>log_{a}y$, then $x<y$.

Precisely, the logarithmic function $f(x)=log_{a}x$ is **monotonically decreasing** for **$0<a<1$**.

**3)** The **argument** of a logarithm must be a **positive** number!

**Logarithmic inequalities – same bases**

**Example 1: **

$$log_{2}(2x+1)>log_{2}(x+3)$$

**Solution:**

The base is $a=2$, which is greater than $1$. This implies $2x+1>x+3$.

$$2x+1>x+3$$

$$2x-x>3-1$$

$$x>2$$

**Check:**

**1) **$2x+1>0 \Rightarrow 2x>-1 \Rightarrow x>-\frac{1}{2}$

**2)** $x+3>0 \Rightarrow x>-3$

We have $x>2$, so both conditions are satisfied. The solution is $x>2$, i.e. $x \in \left<2, +\infty\right>$.

*Example 2: *

$$log_{\frac{1}{2}}(x+3)>log_{\frac{1}{2}}(2x+1)$$

*Solution:*

The base is $a=\frac{1}{2}$, which is less than $0$. This implies $x+3<2x+1$. Notice that we got the same inequality as in the previous example, which means that the procedure and the solution are the same. Therefore, the solution is $x>2$, i.e. $x \in \left<2, +\infty\right>$.

**Logarithmic inequalities – different bases**

**Example 3: **

$$log_{2}(2x^2+2)>log_{4}(x^4)$$

**Solution:**

In this case, the bases are different, but similar. Precisely, we know that $2^2=4$.

$$log_{2}(2x^2+2)>log_{2^2}(x^4)$$

Applying the rule $log_{a^m}y=\frac{1}{m}log_{a}y,$

$$log_{2}(2x^2+2)>\frac{1}{2}log_{2}(x^4)$$

$$log_{2}(2x^2+2)>log_{2}(x^4)^{\frac{1}{2}}$$

$$log_{2}(2x^2+2)>log_{2}x^2$$

The base is $a=2>1$, so we conclude

$$2x^2+2>x^2$$

$$x^2>-2,$$

which is satisfied for every $x \in \mathbf{R}$. Therefore, every $x \in \mathbf{R}$ is the solution.

**Example 4:**

$$log_{7}(x+6)>log_{5}(x+6)$$

**Solution:**

Here we have to use a** base change rule**, i.e. **$log_{a}x=\frac{log_{b}x}{log_{b}a}$**

$$\frac{log(x+6)}{log7}>\frac{log(x+6)}{log5}$$

$$log5 \cdot log(x+6)>log7 \cdot log(x+6)$$

$$(log5-log7) \cdot log(x+6)>0$$

Since $log5-log7$ is a negative number, $log(x+6)$ also must be negative. Otherwise, the product would be negative number. Therefore,

$$log(x+6)<0$$

We”ll use a rule (**logarithm of the base**) **$log_{a}a=1$:**

$$log(x+6)<0 \cdot log10$$

$$log(x+6)<log10^0$$

$$log(x+6)<log1$$

Since the base is $a=10>0$,

$$x+6<1$$

$$x<-5$$

**Check:**

$$x+6>0 \Rightarrow x>-6$$

Since we have $x<-5$, the solution is $-6<x<-5$, i.e. $x \in \left<-6, -5\right>$.

## Other examples

**Example 5: **

$$log_{5}(x+3)>1$$

**Solution:**

In this inequality we have logarithm only on one side. We would like to have a logarithm with same base on both sides. For that purpose, we”ll use a rule (**logarithm of the base**) **$$log_{a}a=1$$**

Now we have

$$log_{5}(x+3)>log_{5}5$$

The base is $a=5$, which is greater than $0$. This implies $x+3>5$. Therefore, $x>2$.

**Check:**

$x+3>0 \Rightarrow x>-3$

We have $x>2$, so the condition is satisfied.

We conclude that $x>2$, i.e. $x \in \left<2, +\infty\right>$ is the solution.

**Example 6: **

$$log_{\frac{1}{2}}(x-4)\geq -2$$

**Solution:**

This inequality is similar to inequality from the previous example; we also have a logarithm only on the one side. We can write the right side as $-2 \cdot 1$ and then we have:

$$log_{\frac{1}{2}}(x-4)\geq -2$$

$$log_{\frac{1}{2}}(x-4)\geq -2 \cdot 1$$

Again, we use a logarithm of the base rule

$$log_{\frac{1}{2}}(x-4)\geq -2 \cdot log_{\frac{1}{2}}\frac{1}{2}$$

Now we can use a **power rule**

$$log_{\frac{1}{2}}(x-4)\geq log_{\frac{1}{2}}\left(\frac{1}{2}\right)^{-2}$$

We now have logarithms on the both sides, which we wanted. The base is $0<a=\frac{1}{2}<1$, which means that

$$x-4\leq \left(\frac{1}{2}\right)^{-2}$$

$$x-4\leq 4$$

$$x\leq 8 $$

**Check:**

$$x-4>0 \Rightarrow x>4$$

We have $x\leq 8 $, so the solution is obviously $4<x \leq 8$, i.e. $x \in \left<4,8\right]$.

**Example 7: **

$$\frac{log^2 x -3 log x+3}{log x-1}<1$$

**Solution:**

We will subtract $1$ from the left side of inequality:

$$\frac{log^2 x -3 log x+3}{log x-1}-1<0$$

Finding the least common denominator, we get

$$\frac{log^2 x -3 log x+3 – log x+1}{log x-1}<0$$

$$\frac{log^2 x -4 log x+4}{log x-1}<0$$

Using a substitution $logx=t$,

$$\frac{t^2-4t+4}{t-1}<0$$

$$\frac{(t-2)^2}{t-1}<0$$

This fraction is obviously a negative number and the value of the numerator is positive number. This means that value of the denominator is negative number, i.e.

$$t-1<0$$

$$t<1,$$

which means

$$logx<1$$

$$logx<log10$$

Since the base is $a=10>1$,

$$x<10.$$

**Check:**

$$x>0$$

Since we have $x<10$, the solution is $0<x<10$, i.e. $x \in \left <0, 10\right>$.