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How to solve the equation $\log_{2x+3}(6x^2+23x+21)=4-\log_{3x+7}(4x^2+12x+9)$ ?

Can someone please tell me a few steps as to how to approach these category of problems? I know $2x+3>0$ and $3x+7>0$ is a must.

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  • $\begingroup$ I would think you could use one of the two: $4 = \log_{2x+3}\left((2x +3)^4\right)$ or $4 = \log_{3x+7}\left((3x+7)^4\right)$. Not sure if that would help or not. $\endgroup$ – Jared Jul 7 '16 at 5:30
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HINT:

Using Laws of Logarithms,

$$\log_{2x+3}(6x^2+23x+21)=\log_{2x+3}(2x+3)+\log_{2x+3}(3x+7)=1+\log_{2x+3}(3x+7)$$

Let $\log_{2x+3}(3x+7)=a$

$$\log_{3x+7}(4x^2+12x+9)=\log_{3x+7}(2x+3)^2=2\log_{3x+7}(2x+3)=\dfrac2a$$ as $\log_{3x+7}(2x+3)\cdot\log_{2x+3}(3x+7)=1$

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Hint: Factorize both polynomials: $$ 6x^2 + 23x + 21= (3x + 7)(2x+3)$$ $$4x^2 + 12x + 9 = (2x+3)^2 $$

After that, work with the properties of logarithms.

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