# Logarithmic equations of second degree

When we have a logarithmic equation of second degree, we need to get rid of the logarithms and obtain an equivalent equation of second degree.

$$\log(x^2+2x)-\log 8=0$$$We can move the constant to the other side of the equation in order to eliminate the logarithms. We obtain a second degree equation that we know how to solve: $$\log(x^2+2x)=\log 8 \Rightarrow x^2+2x=8 \Rightarrow x^2+2x-8=0$$$ To solve it it is necessary to remember the formula: $$\displaystyle x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$$So that: $$\displaystyle x=\frac{-2 \pm \sqrt{4-4\cdot (-8)}}{2}=\frac{-2 \pm \sqrt{4+32}}{2}=\frac{-2\pm \sqrt{36}}{2}=\frac{-2\pm 6}{2}$$$ Therefore, the solutions to the equation will be: $$\displaystyle x=\frac{-2+6}{2}=\frac{4}{2}=2 \\ \displaystyle x=\frac{-2-6}{2}=\frac{-8}{2}=-4$$$But it is necessary to bear in mind that some of the solutions will not be valid for a logarithmic equation, since the logarithm is only defined by numbers greater than $$0$$. So we need to verify that when substituting $$x$$ we obtain negative values. Let's check then: When $$x=2$$ $$x^2+2x \Rightarrow 2^2+2 \cdot 2=4+4=16 >0$$$ Then, the solution is valid.

For $$x =-4$$ $$x^2+2x \Rightarrow (-4)^2+2 \cdot (-4)=16-8=8 >0$$$Therefore, this solution is also valid. In the previous case, we obtained an equation of second degree that we could immediately recognise. Sometimes this might be a bit more tricky. $$\log(9-x^2)=2\log(3x-3)$$$ By the property of the power of logarithms we can obtain: $$\log(9-x^2)=\log(3x-3)^2$$$Thus, the logarithms can be eliminated and we can work with the equations: $$9-x^2=(3x-3)^2 \Rightarrow 9-x^2=9x^2-18x+9 \Rightarrow 9x^2+x^2-18x+9-9=0 \Rightarrow$$$ $$\Rightarrow 10x^2-18x=0$$$Note that we can simplify our equation and we are left with no constant. We can then extract the comon factor $$x$$ and obtain: $$x(10x-18)=0$$$ Thus: $$\begin{array}{l}x=0\\ \\ 10x-18=0 \Rightarrow \displaystyle x=\frac{18}{10}\Rightarrow x=\frac{9}{5} \end{array}$$$And these are the two candidate solutions. We need to verify that they are well defined, that is, that when we put them back into the logarithm we do not get a negative number: When $$x=0$$ $$9-x^2 \Rightarrow 9-0=9>0$$$ and $$(3x-3)^2 \Rightarrow (3\cdot 0-3)^2= 9>0$$$Therefore $$x=0$$ is a solution of the logarithmic equation. When $$x =\displaystyle \frac{9}{5}$$ $$\displaystyle 9-x^2\Rightarrow 9-\Big( \frac{9}{5}\Big)^2=9-\frac{81}{25}>0$$$ and $$\displaystyle (3x-3)^2 \Rightarrow \Big(3\cdot \frac{9}{5}-3\Big)^2=\Big(\frac{27}{5}-3\Big)^2=\Big(\frac{27-5}{5}\Big)^2=\frac{12^2}{5^2}>0$$$Then, it is also a valid solution. Let's see a last example before going on to the exercises: $$\displaystyle \log \sqrt{2x}=\log (x-3)+\log 2$$$ As in the previous cases, it is necessary to try to get rid of the logarithms.

To do this we will use the fact that the sum of the logarithms is the logarithm of its product: $$\displaystyle \log\sqrt{2x}=\log (2 \cdot (x-3))$$$Thus we can eliminate the logarithms to obtain the following equivalent equation: $$\displaystyle \sqrt{2x}=2 \cdot (x-3) \Rightarrow \sqrt{2x}=2x-6$$$ Now it is necessary to get rid of the square root. In order to do so we can square the terms in each of the sides, and we can see that we actually obtain an equation of second degree: $$2x=(2x-6)^2$$$We can extend the righthand side term in order to obtain the second degree equation: $$2x=4x^2-24x+36 \Rightarrow 4x^2-24x-2x+36=0 \Rightarrow 4x-26x+36=0$$$

Note that we can divide all the coefficients by $$2$$ and nothing changes, obtaining a simplified equation: $$\displaystyle \frac{4x^2-26x+36}{2}=0 \Rightarrow 2x^2-13x+18=0$$$We can now use the formula to solve for $$x$$: $$\displaystyle x=\frac{13 \pm \sqrt{13^2-4 \cdot 2 \cdot 18}}{2 \cdot 2}=\frac{13 \pm \sqrt{169-144}}{4}=\frac{13 \pm \sqrt{25}}{4}=\frac{13 \pm 5}{4}$$$

So the possible solutions will be: $$\begin{array}{rcl}x & = &\displaystyle \frac{13+5}{4}=\frac{18}{4}=\frac{9}{2}\\ x&=&\frac{13-5}{4}=\frac{8}{4}=2\end{array}$$$Now it is necessary to verify that the values we found are actually a solution, since we cannot take the logarithm of a negative number. Let's check both solutions: If $$x =\displaystyle \frac{9}{2}$$: $$x-3 \Rightarrow \frac{9}{2} -3 =\frac{9-6}{2}=\frac{3}{2} >0$$$ Therefore, the first value is a result of the logarithmic equation.

If $$x=2$$: $$x-3 \Rightarrow 2-3=-1 < 0$$\$ Note that we obtain a negative number so we have to rule out this solution.

Thus we are left with only one solution, namely: $$x=\displaystyle \frac{9}{2}$$.