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Question

Solve the equation in each of the following.
(i) $$\log _{ 4 }{ \left( x+4 \right)  } +\log _{ 4 }{ 8 } =2$$
(ii) $$\log _{ 6 }{ \left( x+4 \right)  } -\log _{ 6 }{ \left( x-1 \right)  } =1$$
(iii) $$\log _{ 2 }{ x } +\log _{ 4 }{ x } +\log _{ 8 }{ x } =\dfrac { 11 }{ 6 }$$
(iv) $$\log _{ 4 }{ \left( 8\log _{ 2 }{ x }  \right)  } =2$$
(v) $$\log _{ 10 }{ 5 } +\log _{ 10 }{ \left( 5x+1 \right)  } =\log _{ 10 }{ \left( x+5 \right)  } +1$$
(vi) $$4\log _{ 2 }{ x } -\log _{ 2 }{ 5 } =\log _{ 2 }{ 125 } $$
(vii) $$\log _{ 3 }{ 25 } +\log _{ 3 }{ x } =3\log _{ 3 }{ 5 } $$
(viii) $$\log _{ 3 }{ \left( \sqrt { 5x-2 }  \right)  } -\dfrac { 1 }{ 2 } =\log _{ 3 }{ \left( \sqrt { x+4 }  \right)  } $$


Solution

(i) $$\:\:\:\log_{4}{(x+4)}.8=2\Rightarrow \:8(x+4)=4^{2}$$
$$\Rightarrow \: x+4=2\Rightarrow \: x=-2$$

(ii) $$\:\:\:\log_{6} \dfrac{(x+4)}{(x-1)}=1 \Rightarrow \: \dfrac{x+4}{x-1}=6$$
$$\Rightarrow \: x+4=6(x-1)\Rightarrow \:x=2$$

(iii) $$\:\:\:\log_{2}{x}+\log_{2^{2}}{x}+\log_{2^{3}}{x}=\dfrac{11}{6}$$
$$\Rightarrow \: \log_{2}{x}+\dfrac{1}{2}\log_{2}{x}+\dfrac{1}{3}\log_{2}{x}=\dfrac{11}{6}$$
$$\Rightarrow \: \dfrac{11}{6}\log_{2}{x}=\dfrac{11}{6}\Rightarrow \:x=2$$

(iv) $$\:\:\:8\log_{2}{x}=4^{2}=16\Rightarrow \: \log_{2}{x}=2\Rightarrow \: x=4$$

(v) $$\:\:\:\log_{10}{5(5x+1)}=\log_{10}{(x+5).10}$$
$$\Rightarrow \: 5(5x+1)=10(x+5)\Rightarrow \: x=3$$

(vi) $$\:\:\:4\log_{2}{x}=\log_{2}{5^{3}}+\log_{2}{5}=4\log_{2}{5}\Rightarrow \: x=5$$

(vii) $$2\log_{3}{5}+\log_{3}{x}=3\log_{3}{5}\Rightarrow \: x=5$$

(viii) $$\log_{3}\left ( \dfrac{\sqrt{5x-2}}{\sqrt{x+4}} \right )=\dfrac{1}{2}\Rightarrow \: \sqrt{\dfrac{5x-2}{x+4}}=\sqrt{3}\Rightarrow \: x=7$$


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