Answer :
NO 3
[tex]\begin{gathered} 2\log (y+5)=\log 20\text{ -log5} \\ \log (y+5)^2=\log (\frac{20}{5}) \\ (y+5)^2=4 \\ \text{square root both sides} \\ y+5\text{ =2} \\ y=2-5 \\ y=-3 \\ The\text{ right option is b} \end{gathered}[/tex]N0 8
[tex]\begin{gathered} \ln \text{ 8 +ln(n-9)=5ln2} \\ \ln ((n-9)\times8)=\ln 2^5 \\ \ln (8n-72)=\ln 32 \\ 8n-72=32 \\ 8n\text{ = 32+72} \\ 8n=104 \\ n=\frac{104}{8} \\ n=13 \\ \text{Therefore the right answer is f} \end{gathered}[/tex]No 9
[tex]\begin{gathered} 5\ln (2a-1)\text{ =15} \\ \text{Divide both sides by 5} \\ \ln (2a-1)=3 \\ \text{Take the exponential value of both sides} \\ 2a-1=e^3 \\ 2a-1=20.0855 \\ 2a=1+20.0855 \\ 2a=21.0855 \\ \text{Divide both sides by 2} \\ a=\frac{21.0855}{2} \\ a=10.5428 \\ \text{The right option is }q \end{gathered}[/tex]No 11
[tex]\begin{gathered} (\frac{1}{8})^{2-2n}=16 \\ (\frac{1}{2^3})^{2-2n}=2^4 \\ 2^{-3(2-2n)}=2^4 \\ 2^{-6+6n}=2^4 \\ -6+6n\text{ =4} \\ \text{collect like terms} \\ 6n=4+6 \\ 6n\text{ =10} \\ \text{Divide both sides by 6} \\ n=\frac{10}{6} \\ n=\frac{5}{3} \\ \text{Therefore the right answer is }j \end{gathered}[/tex]No 12
[tex]\begin{gathered} -3e^{2m-5}-7=-34 \\ \text{collect like terms} \\ -3e^{2m-5}=-34+7 \\ -3e^{2m-5}=-27 \\ \text{Divide both sides by -3} \\ e^{2m-5}=\frac{-27}{-3} \\ e^{2m-5}=9 \\ \text{Find the natural logarithm of both sides} \\ 2m-5\text{ =ln9} \\ 2m-5=2.1972 \\ 2m=5+2.1972 \\ 2m=7.1972 \\ \text{divide both sides by 2} \\ m=\frac{7.1972}{2} \\ m=\text{ 3.5986} \\ \text{Therefore, the right answer is, p} \end{gathered}[/tex]N0 13
[tex]\begin{gathered} \log _3(p+4)\text{ + }\log _3(p-2)=3 \\ \log _3(p+4)(p-2)\text{ =3} \\ (p+4)(p-2)=3^3 \\ \text{opening up the bracket} \\ p^2-2p+4p-8=27 \\ p^2+2p-8-27=0 \\ p^2+2p-35=0 \\ We\text{ factorize the above equation} \\ p^2+7p-5p-35=0 \\ p(p+7)-5(p+7)=0 \\ (p-5)(p+7)=0 \\ p=+5\text{ or p=-7} \\ \text{Therefore, the right answer is, e} \end{gathered}[/tex]No 14
[tex]\begin{gathered} \ln k-\ln 14=2 \\ \ln (\frac{k}{14})\text{ = 2} \\ Take\text{ the exponential of both sides} \\ \frac{k}{14}=e^2 \\ \frac{k}{14}=7.38906 \\ \text{crossmultiply} \\ k=14\times7.38906 \\ k=103.4468 \\ \text{Therefore the right answer is, u} \end{gathered}[/tex]