Question

Solve the equation in each of the following.
(i) ${\log }_4(x+4)+{\log }_{4}8=2$
(ii) ${\log }_6(x+4)-{\log }_6(x-1)=1$
(iii) ${\log }_{2}x+{\log }_{4}x+{\log }_{8}x=\frac{11}{6}$
(iv) ${\log }_4\left(8{\log }_{2}x\right)=2$
(v) ${\log }_{10}5+{\log }_{10}(5x+1)={\log }_{10}(x+5)+1$
(vi) $4{\log }_{2}x-{\log }_{2}5={\log }_{2}125$
(vii) ${\log }_{3}25+{\log }_{3}x=3{\log }_{3}5$
(viii) ${\log }_3\left(\sqrt{5x-2}\right)-\frac{1}{2}={\log }_3\left(\sqrt{x+4}\right)$

Answer 1

(i) ${\log }_4(x+4).8=2\Rightarrow 8(x+4)=4^2$

$\Rightarrow x+4=2\Rightarrow x=-2$

(ii) ${\log }_6\frac{(x+4)}{(x-1)}=1\Rightarrow \frac{x+4}{x-1}=6$

$\Rightarrow x+4=6(x-1)\Rightarrow x=2$

(iii) ${\log }_{2}x+{\log }_{2^2}x+{\log }_{2^3}x=\frac{11}{6}$

$\Rightarrow {\log }_{2}x+\frac{1}{2}{\log }_{2}x+\frac{1}{3}{\log }_{2}x=\frac{11}{6}$

$\Rightarrow \frac{11}{6}{\log }_{2}x=\frac{11}{6}\Rightarrow x=2$

(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) ${\log }_3\left(\frac{\sqrt{5x-2}}{\sqrt{x+4}}\right)=\frac{1}{2}\Rightarrow \sqrt{\frac{5x-2}{x+4}}=\sqrt{3}\Rightarrow x=7$