Answer: The answer is (C) C. x=1 and x=4.
Step-by-step explanation: We are given to find the true solution of the following equation involving logarithms:
[tex]2\log_36x-\log_34x=2\log_3(x+2).[/tex]
We will be using the following properties of logarithms in the solution.
[tex](i)~a\log b=\log a^b,\\\\(ii)~\log a-\log b=\log\dfrac{a}{b}.[/tex]
The solution is as follows:
[tex]2\log_36x-\log_34x=2\log_3(x+2)\\\\\Rightarrow \log_3(6x)^2-\log_34x=\log_3(x+2)^2\\\\\\\Rightarrow \log_3\dfrac{36x^2}{4x}=\log_3(x^2+4x+4)\\\\\\\Rightarrow 9x=x^2+4x+4\\\\\Rightarrow x^2-5x+4=0\\\\\Rightarrow (x-4)(x-1)=0\\\\\Rightarrow x-4=0,~~~x-1=0\\\\\Rightarrow x=4,~~~x=1.[/tex]
Since we can find the logarithm of a positive integer only, and both the solutions x = 1 and 4 satisfy this condition after substituting in the given equation, so both the solutions are TRUE.
Thus, the correct option is (C).
Answer:
C: x=1 and x=4
Step-by-step explanation:
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