Problems from Logarithmic equations of first degree

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: $$\log ({\displaystyle \frac{x}{3}}+2)-\log (2x-5)=0\Rightarrow \log ({\displaystyle \frac{x}{3}}+2)=\log (2x-5)$$ And it is enough to solve: $${\displaystyle \frac{x}{3}}+2=2x-5\Rightarrow {\displaystyle \frac{x}{3}}-2x=-5-2\Rightarrow {\displaystyle \frac{x-6x}{3}}=7\Rightarrow {\displaystyle \frac{-5x}{3}}=-7\Rightarrow$$ $$\Rightarrow -5x=-21\Rightarrow x={\displaystyle \frac{21}{5}}$$

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: $$\log ({\displaystyle \frac{x+5}{x-4}})=1\Rightarrow {\displaystyle \frac{x+5}{x-4}}={10}^1$$ And it is enough to solve: $$x+5=10\cdot (x-4)\Rightarrow x+5=10x-40\Rightarrow x-10x=-40-5\Rightarrow$$ $$\Rightarrow -9x=-45\Rightarrow x={\displaystyle \frac{-45}{-9}}=5$$

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: $$(x+1)\cdot (x-3)=x^2\Rightarrow x^2-3x+x-3=x^2\Rightarrow x^2-x^2-2x=3\Rightarrow x=-{\displaystyle \frac{3}{2}}$$

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: $$x+1\Rightarrow -{\displaystyle \frac{3}{2}}+1\Rightarrow {\displaystyle \frac{-3+2}{2}}=-{\displaystyle \frac{1}{2}}$$ The result is negative, therefore $-{\displaystyle \frac{3}{2}}$ is not a solution to the equation. So that the last equation has no solution.

Solution:

a) $x={\displaystyle \frac{21}{5}}$

b) $x=5$

c) It has no solution.

Hide solution and development