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

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