Question

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

Answer 1

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

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

Answer 2

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:

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

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

So there is only one solution x = 8.