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Evaluation of $x$ in $$\log_{\frac{3}{4}}\left(\frac{x}{3}\right)+\log_{\frac{1}{2}}\left(\frac{x}{2}\right) = -2$$

$\bf{My\; Try::}$ Here $x>0\;,$ Now Using Properties of $\log\;,$ We get $$\log_{\frac{3}{4}}(x)-\log_{\frac{3}{4}}\left(3\right)+\log_{\frac{1}{2}}(x)+1=-2$$

So $$\log_{\frac{3}{4}}(x)+\log_{\frac{1}{2}}(x)=-3+\log_{\frac{3}{4}}\left(3\right)$$

Now How can I solve after that , Help Required, Thanks

juantheron
  • 53,015

2 Answers2

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HINT:

Using Laws of Logarithms,

$$\log_{\frac34}\dfrac x3=\dfrac{\log_2(x/3)}{\log_2(3/4)}=\dfrac{\log_2x-\log_23}{\log_23-2}$$

and $$\log_{\frac12}\dfrac x2=\dfrac{\log_2(x/2)}{\log_2(2^{-1})}=1-\log_2x$$

lab bhattacharjee
  • 274,582
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$$\frac{\log_2(x)-\log_2(3)}{\log_2(3)-\log_2(4)}+\frac{\log_2(x)-\log_2(2)}{\log_2(1)-\log_2(2)}=-2$$

$$\frac{\log_2(x)-\log_2(3)}{\log_2(3)-2}+\frac{\log_2(x)-1}{0-1}=-2$$

$$\frac{\log_2(x)-\log_2(3)}{\log_2(3)-2}+1-\log_2(x)=-2$$

$$\frac{\log_2(x)-\log_2(3)}{\log_2(3)-2}=-3+\log_2(x)$$

$$\log_2(x)(\frac1{\log_2(3)-2}-1)=-3+\frac{\log_2(3)}{\log_2(3)-2}$$ $$\log_2(x)\frac{3-\log_2(3)}{\log_2(3)-2}=\frac{6-2\log_2(3)}{\log_2(3)-2}$$

$$\log_2(x)=2$$

$$x=4$$

Arashium
  • 2,541