Respuesta :
Answer: The required solution is x = 1.
Step-by-step explanation: We are given to solve the following logarithmic equation :
[tex]\log_5x+\log_5(2x+3)=1~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
We will be using the following logarithmic properties in solving the given equation :
[tex](i)~\log_ab+\log_ac=\log_abc,\\\\(ii)~\log_ab=c~~~~~\Rightarrow b=a^c.[/tex]
From equation (i), we have
[tex]\log_5x+\log_5(2x+3)=1\\\\\Rightarrow \log_5\{x(2x+3)\}=1\\\\\Rightarrow \log_5(2x^2+3x)=1\\\\\Rightarrow 2x^2+3x=5^1\\\\\Rightarrow 2x^2+3x=5\\\\\Rightarrow 2x^2+3x-5=0\\\\\Rightarrow 2x^2+5x-2x-5=0\\\\\Rightarrow x(2x+5)-1(2x+5)=0\\\\\Rightarrow (x-1)(2x+5)=0\\\\\Rightarrow x-1=0,~~~~~2x+5=0\\\\\Rightarrow x=1~~~~~~~\Rightarrow x=-\dfrac{5}{2}.[/tex]
Since the logarithm of a negative number does not exist, so we get
x = 1.
Thus, the required solution is x = 1.