Question

Hello, I need help finding the solutions of the equation log base 2 (x+8)+2=2 log base 2 (x)

Thank you

Answer 1

Domain: x > 0

We know that:
1.

`\log_aa=1` so
`\log_22=1` and


`2=2\cdot 1=2\cdot\log_22`

2.


`c\log_a b = \log_ab^c`

3.


`\log_ax+\log_ay=\log_a(x\cdot y)`

We have equation:


`\log_2(x+8)+2=2\log_2x\qquad\qquad\text{(from 1.)}\\\\ \log_2(x+8)+2\log_22=2\log_2x\qquad\qquad\text{(from 2.)}\\\\ \log_2(x+8)+\log_22^2=\log_2x^2\\\\ \log_2(x+8)+\log_24=\log_2x^2\qquad\qquad\text{(from 3.)}\\\\ \log_2(x+8)\cdot4=\log_2x^2\\\\ (x+8)\cdot4=x^2\\\\4x+32=x^2\\\\x^2-4x-32=0`


`a=1\qquad\qquad b=-4\qquad\qquad c=-32\\\\\\\Delta=b^2-4ac=(-4)^2-4\cdot1\cdot(-32)=16+128=144\\\\√(\Delta)=√(144)=12\\\\\\x_1=(-b-√(\Delta))/(2a)=(-(-4)-12)/(2)=(4-12)/(2)=(-8)/(2)=-4\ \textless \ 0\,\,\text{not a solution}}\\\\\\ x_2=(-b+√(\Delta))/(2a)=(-(-4)+12)/(2)=(4+12)/(2)=(16)/(2)=\boxed{8}\ \textgreater \ 0`

So there is only one solution x = 8.

Answer 2

`\bf log_{{ a}}\left( (x)/(y)\right)\implies log_{{ a}}(x)-log_{{ a}}(y) \\\\\\ % Logarithm of exponentials log_{{ a}}\left( x^{{ b}} \right)\implies {{ b}}\cdot log_{{ a}}(x) \\\\\\ {{ a}}^{log_{{ a}}x}=x\impliedby \textit{log cancellation rule}\\\\ -----------------------------\\\\`


`\bf log_2(x+8)+2=2log_2(x)\implies log_2(x+8)-2log_2(x)=-2 \\\\\\ log_2\left(\cfrac{x+8}{x^2} \right)=-2\implies 2^{log_2\left(\cfrac{}{}(x+8)/(x^2) \right)}=2^(-2)\implies \cfrac{x+8}{x^2}=\cfrac{1}{2^2} \\\\\\ 4x+32=x^2\implies 0=x^2-4x-32 \\\\\\ 0=(x+4)(x-8)\implies \begin{cases} 0=x+4\implies &-4=x\\ 0=x-8\implies &8=x \end{cases}`