**Introduction**

We’ve learned that the expression $log_{a}y=x$ is equivalent to expression $a^x=y$, where $a>0, a \neq 1$.

For example, $log_{5}25=2$, since $5^2=25$. But, what if we don’t know the exact value of $y$? Let’s say, how to solve an equation $log_{2}4x+7=5$? For that purpose, we need to introduce **logarithmic equations**.

Let’s solve mentioned example.

**Example 1: **

$$log_{2}4x+7=5$$

**Solution:**

Simply, we’ll just use the well – known fact: $log_{a}y=x \Leftrightarrow a^x=y$.

Now we have: $$4x + 7 = 2^5$$

Obviously, we’ve got linear equation in one variable, which we need to solve for $x$.

$$4x + 7 = 32$$

$$4x = 32 – 7$$

$$4x = 25$$

$$x = \frac{25}{4} $$

We mustn’t forget about **conditions**! I.e., logarithms are defined only for positive numbers, so we need to make sure that the value of the expression $4x + 7$ is $4x + 7 > 0$. Let’s check that.

**Check: **

$$4x + 7 > 0$$

$$4 \cdot \frac{25}{4} + 7 =32 >0$$

Therefore, we conclude that $x= \frac{25}{4}$ is really the solution of given logarithmic equation.

**Examples**

The main idea of logarithmic equations is to get equation $log_{a}x=log_{a}y$, and then we can conclude $x = y$.

* Example 2: *$$log (x-1) + log (x-2) = 2 log (x-3)$$

**Solution: **

When we see the summation of logarithms, the first thinking that comes up to mind is to use the product rule.

$$log [(x-1)\cdot(x-2)]=2 log (x-3)$$

But, we would like to have an equation $log_{a}x=log_{a}y$, so number $2$ ”bothers” us. We”ll just apply the power rule:

$$log [(x-1)\cdot(x-2)]=log (x-3)^2$$

Now we have

$$(x-1)\cdot(x-2)= (x-3)^2$$

$$x^2-2x-x+2=x^2-6x+9$$

$$x^2-3x+2=x^2-6x+9$$

$$3x=7$$

$$x=\frac{7}{3}$$

**Check:**

Now we have several conditions:

**1)** $x-1 >0$

$\frac{7}{3} -1 = \frac{4}{3} >0 $

**2)** $x-2>0$

$\frac{7}{3} – 2 = \frac{1}{3} >0$

**3) **$x-3>0$

$\frac{7}{3} – 3 = – \frac{2}{3}<0$

Condition **3)** is not valid, so $x=\frac{7}{3}$ is not the solution. This equation has no solutions.

**Example 3: **

$$log (3x-5)-\frac{1}{2}log(x+1)=1-log5$$

**Solution:**

When we see the subtraction of logarithms, the first thinking that comes up to mind is to use the quotient rule. But first, we”ll apply the power rule:

$$log (3x-5)-log(x+1)^\frac{1}{2}=1-log5$$

$$log \frac {3x-5}{(x+1)^\frac{1}{2}}=1-log5$$

$$log \frac {3x-5}{\sqrt{x+1}}=1-log5$$

Since we want $log_{a}b=log_{a}c$, we need to rearrange the right side of the equation. Remember that $1=log10$:

$$log \frac {3x-5}{\sqrt{x+1}}=log10-log5$$

Now use a quotient rule for the right side of the equation:

$$log \frac {3x-5}{\sqrt{x+1}}=log\frac{10}{5}$$

$$log \frac {3x-5}{\sqrt{x+1}}=log2$$

We conclude: $$\frac {3x-5}{\sqrt{x+1}}=2.$$

Now we have

$$3x-5=2 \cdot \sqrt{x+1}$$

By squaring, we get

$$(3x-5)^2=4 \cdot (x+1) $$

$$9x^2-30x+25=4x+4$$

$$9x^2-34x+21=0,$$

which is a quadratic equation.

$$x_{1,2}=\frac{34\pm \sqrt {34^2-4\cdot 9\cdot 21}}{2 \cdot 9}=\frac{34 \pm \sqrt{1156-36 \cdot 21}}{18}=\frac{34 \pm 20}{18}$$

$$x_{1}=3, x_{2}=\frac{7}{9}$$

Obviously, we have two possible solutions.

**Check:**

**1)** $3x – 5 > 0$

For $x_{1}=3$: For $x_{2}=\frac{7}{9}$:

$3 \cdot 3 – 5 = 4 >0$ $3 \cdot \frac{7}{9} – 5 = – \frac{8}{3} <0 $

We immediately conclude that $x_{2}=\frac{7}{9}$ is not the solution, since both conditions must be satisfied.

**2)** $x + 1 > 0$

For $x_{1}=3$:

$3 + 1 = 4 > 0$

The solution of the equation is $x=3$.

**Example 4: **

$$\frac{1}{5-4logx}+\frac{4}{1+logx}=3$$

**Solution:**

For simplicity, let’s use substitution: $logx=t$. Now we have

$$\frac{1}{5-4t}+\frac{4}{1+t}=3$$

Multiplying the whole equation by $(5-4t)\cdot (1+t)$, we get

$$1 + t + 4 \cdot (5 – 4t)=3 \cdot (5 – 4t)(1 + t)$$

$$1 + t + 20 – 16t = 3 \cdot (5 + 5t -4t -4t^2)$$

By rearranging, we get quadratic equation

$$2t^2 – 3t + 1 = 0$$

$$t_{1,2}=\frac{3 \pm \sqrt{9 – 8}}{4}$$

$$t_{1}=1, t_{2}=\frac{1}{2}$$

and, since we used substitution $logx=t$, there are two possible cases:

$$1°) logx = 1$$

$$2°) logx = \frac{1}{2}$$

1°) $logx=1$ 2°) $logx = \frac{1}{2}$

$x = 10$ $x = 10^\frac{1}{2}$

$x = \sqrt{10}$

**Check:**

**1) **$x > 0$

For $x = 10$ : For $x = \sqrt{10}$:

$10 > 0$ $\sqrt{10} >0$

When multiplying by $(5-4t)\cdot (1+t)$, we need to make sure that both factors are $\neq 0$:

**2)** $5 – 4t \neq 0$

For $t = 1$ : For $t = \frac{1}{2}$:

$5 – 4 \cdot 1 = 1 \neq 0$ $5 – 4 \cdot \frac{1}{2} = 3 \neq 0$

**3)** $1 + t \neq 0$

For $t = 1$ : For $t = \frac{1}{2}$:

$1 + 1 = 2 \neq 0$ $1 + \frac{1}{2} = \frac{3}{2} \neq 0 $

All conditions are fulfilled, so the solutions are $x_{1} = 10$ and $x_{2} = \sqrt{10}$.

**Example 5: **

$$log_{3}x+log_{9}x+log_{27}x= \frac{11}{12}$$

**Solution:**

We”ll apply the rule: **$$log_{a^m}y=\frac{1}{m}log_{a}y.$$**

Now we have

$$log_{3}x+log_{3^2}x+log_{3^3}x= \frac{11}{12}$$

$$log_{3}x+\frac{1}{2}log_{3}x+\frac{1}{3}log_{3}x= \frac{11}{12}$$

Multiplying the whole equation by $12$, we get

$$12log_{3}x+6log_{3}x+4log_{3}x=11$$

$$22log_{3}x=11$$

Dividing the whole equation by $22$, we get

$$log_{3}x= \frac{1}{2}$$

$$x=3^\frac{1}{2}=\sqrt{3}$$

**Check:**

$$x>0$$

$$\sqrt{3}>0$$

The solution of the equation is $x=\sqrt{3}$.