# Problems from Logarithmic equations of second degree

Solve the following logarithmic equations:

a) $2\log (x+5)=\log(21-3x^2)$

b) $\log(17-4x)=2\log(2x-1)$

c) $\log x+\log 3=\dfrac{\log 4}{2\log x}$

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### Development:

a) In the first case, we need to use the property by which we can bring the $2$ inside the logarithm by taking the logarithm of the squared expression: $$2\log (x+5)=\log(21-3x^2) \Rightarrow \log (x+5)^2=\log(21-3x^2)$$ At this point, we can eliminate the logarithm on both sides of the identity, so to obtain an equation of second degree that we will know how to solve: $$(x+5)^2=21-3x^2 \Rightarrow x^2+10x+25=21-3x^2 \Rightarrow x^2+3x^2+10x+25-21=0 \Rightarrow$$ $$\Rightarrow 4x^2+10x+4=0$$ To find $x$, it is necessary to apply the formula: $$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-10\pm\sqrt{10^2-4\cdot4\cdot4}}{2\cdot4}=\dfrac{-10\pm\sqrt{36}}{8}=\dfrac{-10\pm6}{8}$$ So the solutions to the equation will be: $$x=\dfrac{-10+6}{8}=\dfrac{-4}{8}=-\dfrac{1}{4} \\ x=\dfrac{-10-6}{8}=\dfrac{-16}{8}=-2$$ Now we need to verify that these are solutions to the logarithmic equation, since the logarithm can only take positive numbers: If $x =-\dfrac{1}{4}$: $$(x+5)^2 \Rightarrow \Big(-\dfrac{1}{4}+5\Big)^2 >0$$ $$(21-3x^2) \Rightarrow \Big(21-3\cdot\Big(-\dfrac{1}{4}\Big)^2\Big) \Rightarrow \Big(21-3\cdot\Big(\dfrac{1}{16}\Big)\Big) \Rightarrow \Big(21-\dfrac{3}{16}\Big) >0$$ So $x =-\dfrac{1}{4}$ is a solution to the equation.

If $x=-2$: $$(x+5)^2 \Rightarrow \Big(-2+5\Big)^2 >0$$ $$(21-3x^2) \Rightarrow (21-3\cdot(-2)^2) \Rightarrow (21-3\cdot4) \Rightarrow (21-12)=9 >0$$ Therefore $x=-2$ is also a solution of the equation.

b) The second case is similar to the first one: $$\log(17-4x)=2\log(2x-1) \Rightarrow \log(17-4x)=\log(2x-1)^2$$ We can now eliminate the logarithm on both sides of the identity $$17-4x=(2x-1)^2 \Rightarrow 17-4x=4x^2-4x+1 \Rightarrow 4x^2-4x+4x+1-17=0 \Rightarrow$$ $$\Rightarrow 4x^2-16=0$$ We solve this second degree incomplete equation $$4x^2=16 \Rightarrow x^2=\dfrac{16}{4} \Rightarrow x^2=4 \Rightarrow x=\pm\sqrt{4}=\pm2$$

In order to be a solution it is necessary to get it greater than 0.

If $x=2$: $$17-4x \Rightarrow 17-4\cdot2=17-8=9 > 0$$ Therefore $x=2$ is a solution of the equation.

If $x=-2$: $$17-4x \Rightarrow 17-4\cdot(-2)=17+8=25 > 0$$ So $x=-2$ is also a solution of the equation.

c) In the last case it is necessary to apply several of the properties of the logarithms to get rid of them and obtain an equivalent equation of the second degree: $$\log x+\log 3=\dfrac{\log 4}{2\log x} \Rightarrow \log 3x=\dfrac{\log 4}{\log x^2} \Rightarrow \log 3x=\log(4-x^2)$$ We can now get rid of the logarithms and obtain an equation of degree two: $$3x=4-x^2 \Rightarrow -x^2-3x+4=0$$ So that: $$x=\dfrac{3\pm\sqrt{3^2-4\cdot(-1)\cdot4}}{2\cdot(-1)}=\dfrac{3\pm\sqrt{9+16}}{-2}=\dfrac{3\pm\sqrt{25}}{-2}=\dfrac{3\pm5}{-2}$$ Therefore the possible solutions will be: $$x=\dfrac{3+5}{-2}=-\dfrac{8}{2}=-4 \\ x=\dfrac{3-5}{-2}=\dfrac{-2}{-2}=1$$ We need to verify that those are indeed solutions to the original equation. Note that $x =-4$ cannot be a solution, since, when replacing it into the logarithm we have: $$\log x \Rightarrow \log(-4)$$ So $x=-4$ is not a solution.

We need to do the same to check that $x=1$ is indeed a solution.

### Solution:

a) $x=-\dfrac{1}{4}; \ x=-2$

b) $x=2; \ x=-2$

c) $x=1$

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