Question

02) Resolva as seguintes equações logarítmicas. > Er dc ساq a) log2 (x+2) = 5 b) 4log: (x – 40) = 16 c) 12 – log, (x - 1) = 10 d) 2log(x - 2)+1=7 ...
na imagem fica melhor. E se possível me ensine essa propriedade de soma por favor

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Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\mathsf{log_2\:(x + 2) = 5}$

$\mathsf{x + 2 = 2^5}$

$\mathsf{x + 2 = 32}$

$\boxed{\boxed{\mathsf{x = 30}}}$

$\mathsf{4\:log_3\:(x - 40) = 16}$

$\mathsf{log_3\:(x - 40) = 4}$

$\mathsf{x - 40 = 3^4}$

$\mathsf{x - 40 = 81}$

$\boxed{\boxed{\mathsf{x = 121}}}$

$\mathsf{12 - \:log_7\:(x - 1) = 10}$

$\mathsf{log_7\:(x - 1) = 2}$

$\mathsf{x - 1 = 7^2}$

$\mathsf{x - 1 = 49}$

$\boxed{\boxed{\mathsf{x = 50}}}$

$\mathsf{2\:log_4\:(x - 2) + 1 = 7}$

$\mathsf{2\:log_4\:(x - 2) = 6}$

$\mathsf{log_4\:(x - 2) = 3}$

$\mathsf{x - 2 = 4^3}$

$\mathsf{x - 2 = 64}$

$\boxed{\boxed{\mathsf{x = 66}}}$

Answer 2

$\Large\boxed{\begin{array}{l}\rm a)~\sf \log_2(x+2)=5\\\sf x+2)=2^5\\\sf x+2=32\\\sf x=32-2\\\sf x=30\\\rm b)~\sf 4\log_3(x-40)=16\\\sf \log_3(x-40)=\dfrac{16}{4}\\\\\sf \log_3(x-40)=4\\\sf x-40=3^4\\\sf x-40=81\\\sf x=40+81\\\sf x=121\end{array}}$

$\Large\boxed{\begin{array}{l}\rm c)~\sf12-\log_7(x-1)=10\\\sf \log_7(x-1)=12-10\\\sf \log_7(x-1)=2\\\sf x-1=7^2\\\sf x-1=49\\\sf x=49+1\\\sf x=50\end{array}}$

$\Large\boxed{\begin{array}{l}\rm d)~\sf2\log_4(x-2)+1=7\\\sf 2\log_4(x-2)=7-1\\\sf 2\log_4(x-2)=6\\\sf \log_4(x-2)=\dfrac{6}{2}\\\\\sf \log_4(x-2)=3\\\sf x-2=4^3\\\sf x-2=64\\\sf x=64+2\\\sf x=66\end{array}}$