Question

Resolva as equações logarítmicas.

a) $log_{4} (x+5) = 2$

b) $log_{2} (x+1) = log_{4}( x^{2} +3)$

c) $log_{x}4^{2} - log_{x} 16 = -1$

Answer 1

a) Apenas usar igualdade do logaritmo.
$\log_a b=c \rightarrow a^c=b\\\\ \log_4(x+5)=2\\ 4^2=x+5\\ x=16-5\\ x=11$

b) Precisamos igualar as bases dos logaritmos para podermos cortá-los.
$\log_2(x+1)=\log_4(x^2+3) \rightarrow 4 = 2^2\\ \log_2(x+1)=\dfrac{1}{2}\log_2(x^2+3)\\ 2*\log_2(x+1)=\log_2(x^2+3)\\ \not \log_2(x+1)^2=\not \log_2(x^2+3)\\ (x+1)^2=x^2+3\\ \not x^2+2x+1=\not x^2+3\\ 2x+1=3\\ x=1$

c) Propriedade da subtração de logaritmos.
$\log_a \frac{b}{c}= \log_ab-\log_ac\\ \\ \log_x4^2-\log_x16=-1\\ \log_x\dfrac{16}{16}=-1\\ x^{-1}=1 \rightarrow x=1\\ Mas~x \neq 1~pois~e~a~base~do~logaritmo.\\ Logo~nao~existe~solucao~para~a~equacao.$

Answer 2

a)

$\mathsf{C.E:\,x>-5}$

$\mathsf{x+5={4}^{2}}\\\mathsf{x+5=16}\\\mathsf{x=16-5}$

$\boxed{\boxed{\mathsf{x=11}}}$

b)

$\mathsf{C.E:\,x+1>0\to\,x>-1}$

$log_{2} (x+1) = log_{4}( x^{2} +3)</p><p>$

$log_{2} (x+1) = log_{2}{( x^{2} +3)}^{\frac{1}{2}} </p><p>$

$log_{2} (x+1) = log_{2}{( x^{2} +3)}^{\frac{1}{2}} \\ \sqrt{ {x}^{2} + 3 } = x + 1 \\ \cancel{x}^{2} + 3 = \cancel{{x}^{2}} + 2x + 1</p><p> \\ 2x = 3 - 1$

$2x = 2 \\ x = \frac{2}{2} = 1$

$\mathsf{s=\{1\}}$

c) $\mathsf{C.<u>E</u><u>:</u><u>\</u><u>,</u>x\ne1}$

$log_{x}4^{2} - log_{x} 16 = -1 \\ log_{x}( \frac{16}{16} ) = - 1 \\ log_{x}(1) = - 1 \\ {x}^{ - 1} = 1 \\ x = 1$

$\mathsf{s=\varnothing}$