Question

Question 32, it’s one am for me and I have exams tomorrow, please be quick and include answers in bold, thanks

Answer 1

The given equation is

`\log _3x+\log _3(2x+1)=1`

Use the rule

`\log _ba+\log _bc=\log _bac`

`\log _3x(2x+1)=1`

Change it to the exponent using the rule

`\log _ba=n\rightarrow b^n=a`

`\begin{gathered} x(2x+1)=3^1 \\ 2x^2+x=3 \end{gathered}`

Subtract 3 from both sides

`\begin{gathered} 2x^2+x-3=3-3 \\ 2x^2+x-3=0 \end{gathered}`

Factorize it

`(2x+3)(x-1)=0`

Equate each bracket by 0 to find x

`\begin{gathered} 2x+3=0 \\ 2x+3-3=0-3 \\ 2x=-3 \\ (2x)/(2)=(-3)/(2) \\ x=-1.5 \end{gathered}`

`\begin{gathered} x-1=0 \\ x-1+1=0+1 \\ x=1 \end{gathered}`

We will refuse x = -1.5 because we can not use negative numbersThank

with log

The answer is x = 1 only

The answer is D