How to solving logarithm equation of $\log_{x}{\left(1+\frac{15}{x}\right)}=2\left(\log_{x}{(10)} -1\right) $

Question

$$\log_{x}{\left(1+\frac{15}{x}\right)}=2\left(\log_{x}{(10)} -1\right) $$ My work so far is :

Step 1 : I searching the definition term of $\log_{x}{\left(1+\frac{15}{x}\right)} $. For x as its base, $$x \neq 1$$ For x as its numerus, $$1+\frac{15}{x} > 0$$ $$\frac{x+15}{x}>0 $$ $$(x+15)(x)>0$$ $$x<-15 \lor x>0 $$

Step 2: I trying simplify the right hand side. $$2(\log_{x}{(10)}-1)=2\log_{x}{(10)} - 2 $$ $$=\log_{x}{(100)}-2 $$

Step 3: Rewrite the equation. $$\log_{x}{\left(1+\frac{15}{x}\right)}=\log_{x}{(100)}-2$$ $$\log_{x}{\left(\frac{x+15}{x}\right)}-\log_{x}{(100)}+2=0$$ What is the next 3 step to solving this equation ? I really thank you for your information, suggestion and answer.

Answer 1

I would write $$\log_{x}\frac{1+\frac{15}{x}}{100}=-2$$ and this is $$\frac{1}{x^2}=\frac{1+\frac{15}{x}}{100}$$ This is $$0=x^2+15x-100$$ Can you solve this equation?

Answer 2

Use that

$$\log_{x}{\left(1+\frac{15}{x}\right)}=2\left(\log_{x}{(10)} -1\right)=2\left(\log_{x}{(10)} -\log_x x\right)=\log_x\left(\frac{10}x\right)^2$$

therefore since $\log$ function is monotonic, providing that $x>0$ and $x\neq 1$, the given identity is equivalent to

$$1+\frac{15}{x}=\left(\frac{10}x\right)^2 \iff x^2+15x-100=0$$

Answer 3

Apart from setting $$x^2+15x=100=0$ one has to also impose the condition at the out set that $x>0$ and (1+15/x)>0$, this will restrict the root to be only one which is $x=5$.

Answer 4

Using logarithm properties, we get $\log_x\left(1+\dfrac{15}{x}\right)=\log_x\left(\dfrac{100}{x^2}\right)\Rightarrow1+\dfrac{15}{x}=\dfrac{100}{x^2}\Rightarrow x^2+15x-100=0\Rightarrow(x+20)(x-5)=0\Rightarrow x=5$ (since $x$ cannot be negative; it is a logarithm base). One can see this simple fact by using the change of base rule and the fact that logarithms only take in positive values.