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If $2\log_3(x-2y)=\log_3x+\log_3y$, find $\frac xy$.

My try:

$$2\log_3(x-2y)=\log_3x+\log_3y$$Combining the terms on the RHS and bringing them over to the LHS $$2\log_3(x-2y)-\log_3xy=0$$ Bring everything into a single $\log$ $$\log_3\frac{(x-2y)^2}{xy}=0$$ Converting this into a solvable quadratic $$x^2-5xy+4y^2=0$$ Factoring, I get $$(x-4)(x-y)=0$$ so $$x=y\;\text{or}\;x=4y$$ Therefore my final answer is $$\frac xy=1\;\text{or}\;\frac xy=4$$

Is my solution correct?

  • 2
    It is fine to me ! – 2017-02-17
  • 1
    There's a typo in the factoring – 2017-02-17

2 Answers 2

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The final answer is not correct.

We have $\log_3(x-2y),\log_3 x$ and $\log_3 y$.

So, we have to have $$x-2y\gt 0\quad\text{and}\quad x\gt 0\quad\text{and}\quad y\gt 0$$ from which we have to have $$\frac xy\gt 2$$

Hence, from what you've done, the answer is $$\color{red}{\frac xy=4}$$

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$$2\log_3(x-2y)=\log_3x+\log_3y$$ $$\log_3(x-2y)^2=\log_3xy$$ $$(x-2y)^2=xy$$ $$x^2-4xy+4y^2-xy=0$$ $$x^2-5xy+4y^2=0$$ $$\frac{x^2}{y^2}-5\frac{x}{y}+4=0$$ $x/y=t$ $$t^2-5t+4=0$$ $$t_{1,2}=\frac{5\pm 3}{2}$$ $$x/y=4,x/y=1$$