Solve the logarithmic equations:
a) $$\log\Big(\dfrac{x}{3}+2\Big)-\log(2x-5)=0$$
b) $$\log(x+5)-1=\log(x-4)$$
c) $$\log(x+1)+\log(x-3)=\log x^2$$
Development:
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.
Solution:
a) $$x=\dfrac{21}{5}$$
b) $$x=5$$
c) It has no solution.