Question

NO LINKS!!!
Please help me #49 and 50

Answer 1

49. a) A - C - 1

49. b) 1 - 2R

49. c) 2Q - P

50. x = 103

Explanation:

`\boxed{\begin{minipage}{5cm}\underline{Log laws}\\\\$\log_axy=\log_ax + \log_ay$\\\\$\log_a \left((x)/(y)\right)=\log_ax - \log_ay$\\\\$\log_ax^n=n\log_ax$\\\\$\log_ab=c \iff b=a^c$\\\\&\log_aa=1$\\ \end{minipage}}`

Question 49

Part a

Given:

`\log_5 9=A`

`\log_5 6=B`

`\log_5 11=C`

`\begin{aligned}\log_5 (9)/(55)&=\log_5 9 - \log_5 55\\&=\log_5 9 - \log_5 (11 \cdot 5)\\&=\log_5 9 - (\log_5 11 + \log_5 5)\\&=\log_5 9 - \log_5 11 - \log_5 5\\&=A-C-1\\\end{aligned}`

Part b

Given:

`\log_3 10=R`

`\log_3 4=S`

`\log_3 11=T`

`\begin{aligned}\log_3 (3)/(100)&=\log_3 3 - \log_3 100\\&=\log_3 3 - \log_3 (10 \cdot 10)\\&=\log_3 3 - \log_3 10^2\\&=\log_3 3 - 2\log_3 10\\&=1-2R\\\end{aligned}`

Part c

Given:

`\log_3 4=P`

`\log_3 10=Q`

`\log_3 7=R`

`\begin{aligned}\log_3 25&=\log_3 (100)/(4)\\&=\log_3 100 - \log_3 4\\&=\log_3 (10 \cdot 10) - \log_3 4\\&=\log_3 10^2 - \log_3 4\\&=2 \log_3 10 - \log_3 4\\&=2Q-P\end{aligned}`

Question 50

The error was in line 4 of the calculation. The log law

• 
  `\log_ab=c \iff b=a^c`

was not applied correctly.

The correct solution is:

`\begin{aligned}3 \log (x-3) + 1 & = 7\\3 \log (x-3) & = 6\\\log (x-3) & = 2\\x-3&=10^2\\x&=3+10^2\\x&=103\end{aligned}`