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 $x=\sqrt{3}$.
