Respuesta :

semsee45

Answer:

x = 0

Step-by-step explanation:

[tex]\large \boxed{\begin{minipage}{8 cm}\underline{Log Laws} \\ \\$\textsf{Product law}: \quad \log_ax + \log_ay=\log_a(xy) \\ \\\textsf{If }\: \log_ab=c\: \textsf{ then }\: a^c=b$\\\end{minipage}}[/tex]

[tex]\large \begin{aligned}\log_9(x+3)+\log_9(x+1) & = \dfrac{1}{2}\\\\\log_9[(x+3)(x+1)] & = \dfrac{1}{2}\\\\9^{\frac{1}{2}} & =(x+3)(x+1)\\\\3 & = x^2+4x+3\\\\x^2+4x & = 0\\\\x(x+4) & = 0 \\\\\implies x & = 0, -4\end{aligned}[/tex]

Cannot take logs of negative numbers, therefore x = 0  only.

[tex]\\ \rm\Rrightarrow log_9(x+3)+log_9(x+1)=\dfrac{1}{2}[/tex]

[tex]\\ \rm\Rrightarrow log_9(x+3)(x+1)=\dfrac{1}{2}[/tex]

[tex]\\ \rm\Rrightarrow (x+3)(x+1)=9^{\frac{1}{2}}[/tex]

[tex]\\ \rm\Rrightarrow (x+3)(x+1)=3[/tex]

[tex]\\ \rm\Rrightarrow x(x+1)+3(x+1)=3[/tex]

[tex]\\ \rm\Rrightarrow x^2+x+3x+3=3[/tex]

[tex]\\ \rm\Rrightarrow x^2+4x=0[/tex]

[tex]\\ \rm\Rrightarrow x(x+4)=0[/tex]

  • x=0,-4

x can't be negative interms of logarithm .

  • x=0

Some logarithm concepts

[tex]\boxed{\boxed{\begin{minipage}{5cm}\displaystyle\circ\sf\ ^{a} log \ a= 1\\\\\circ \ ^{a}log \ 1 = 0 \\\\\circ \ ^{a ^{n}} log \ b^{m}= \dfrac{m}{n} \times\:^{a}log \ b \\\\\circ \ ^{a^{m}} \ log \ b^{m} = \ ^{a}log \ b \\\\\circ \ ^{a}log \ b = \dfrac{1}{^{b}log \ a} \\\\\circ \ ^{a}log \ b = \dfrac{^{m}log \ b}{^{m} log \ a} \\\\\circ \ a^{^{a} logb} = b \\\\\circ \ ^{a}log \ b + ^{a}log \ c = \ ^{a}log(bc) \\\\\circ \ ^{a}log \ b -\: ^{a}log \ c = \ ^{a}log \left( \dfrac{b}{c} \right) \\\circ \ ^{a}log \ b \:\cdot\: ^{a}log \ c = \ ^{a}log \ c \\\\\circ \ ^{a}log \left( \dfrac{b}{c} \right) = \ ^{a}log \left(\dfrac{c}{b}\right)\end{minipage}}}[/tex]