Logarithmic equations of first degree

a) The first equation is simple. We move to the other side of the identity one of the terms. Then we take the logarithms out and we solve the linear equation: $$\log\Big(\dfrac{x}{3}+2\Big)-\log(2x-5)=0 \Rightarrow \log\Big(\dfrac{x}{3}+2\Big)=\log(2x-5)$$ And it is enough to solve: $$\dfrac{x}{3}+2=2x-5 \Rightarrow \dfrac{x}{3}-2x=-5-2 \Rightarrow \dfrac{x-6x}{3}=7 \Rightarrow \dfrac{-5x}{3}=-7 \Rightarrow $$ $$\Rightarrow -5x=-21 \Rightarrow x=\dfrac{21}{5}$$

b) In this case we can put together all the terms containing the variable $x$ on one side, an all the other terms on the other: $$\log(x+5)-1=\log(x-4) \Rightarrow \log(x+5)-\log(x-4)=1$$ Now, we can use the rules for the logarithms (namely, the subtraction of logarithms is the logarithm of the difference) to obtain: $$\log\Big(\dfrac{x+5}{x-4}\Big)=1 \Rightarrow \dfrac{x+5}{x-4}=10^1$$ And it is enough to solve: $$x+5=10\cdot(x-4) \Rightarrow x+5=10x-40 \Rightarrow x-10x=-40-5 \Rightarrow$$ $$\Rightarrow -9x=-45 \Rightarrow x=\dfrac{-45}{-9}=5$$

c) Finally, the third equation seems to be of the second degree, but if we remember the rules of the logarithms we know that the exponent of $x$ can be brought outside of the logarithm.

The first thing that can be done is to group the first terms using the property of the product of logarithms: $$\log(x+1)+\log(x-3)=\log x^2 \Rightarrow \log[(x+1)\cdot(x-3)]=\log x^2$$ At this point, the logarithms can be eliminated to obtain an equivalent equation: $$(x+1)\cdot(x-3)=x^2 \Rightarrow x^2-3x+x-3=x^2 \Rightarrow x^2-x^2-2x=3 \Rightarrow x=-\dfrac{3}{2}$$

But remember that, in order to obtain a valid solution we need a positive number inside the logarithm. So we need to make sure that $x+1$ is indeed positive: $$x+1 \Rightarrow -\dfrac{3}{2}+1 \Rightarrow \dfrac{-3+2}{2}=-\dfrac{1}{2}$$ The result is negative, therefore $-\dfrac{3}{2}$ is not a solution to the equation. So that the last equation has no solution.