Question

Please solve the first one and the last one i boxed

Answer 1

For the first one:

`\begin{gathered} 28-\log _2(x-3)=9 \\ \rightarrow28-9=\log _2(x-3)\rightarrow19=\log _2(x-3) \end{gathered}`

Remember that:

`\log _a(b)=c\Leftrightarrow a^c=b`

This way,

`\begin{gathered} 19=\log _2(x-3)\leftrightarrow2^(19)=x-3\rightarrow2^(19)+3=x \\ \rightarrow x=524291 \end{gathered}`

For the last one:

`\begin{gathered} 200000=100000e^(0.08t) \\ \rightarrow\ln (200000)=\ln (100000e^(0.08t)) \\ \rightarrow\ln (200000)=\ln (100000)+\ln (e^(0.08t)) \\ \rightarrow\ln (200000)-\ln (100000)=(0.08t)\ln (e^{}) \\ \rightarrow\ln (200000)-\ln (100000)=0.08t \\ \rightarrow(\ln (200000)-\ln (100000))/(0.08)=t \\ \rightarrow t=8.66 \end{gathered}`