Question

1- Resolva a equação 2( − 4) = 3

a) 8

b) 9

c)10

d) 11

e) 12

2) Qual é o conjunto verdade de (3 + 4) = (4 + 2) ?

a) 0

b) 1

c) 2

d) 3

e) 4

3) (UP). A solução da equação logarítmica 10( − 4) = 2 é:

a) x = 6

b) x = 10

c) x = 50

d) x = 100

e) x = 104

4) Resolva a equação logarítmica +3

(5x − 1) = 1

a) 0

b) 1

c) 2

d) 3​

Answer 1

$\large\boxed{\begin{array}{l}\rm 1)~Resolva~a~equac_{\!\!,}\tilde ao~\ell og_2(x-4)=3\\\tt a)~\sf8\\\tt b)~\sf9\\\tt c)~\sf10\\\tt d)~\sf11\\\tt e)~\sf12\\\underline{\rm soluc_{\!\!,}\tilde ao\!:}\\\sf\ell og_2(x-4)=3\\\sf x-4=2^3\\\sf x-4=8\\\sf x=8+4\\\sf x=12\\\huge\boxed{\boxed{\boxed{\boxed{\sf\maltese~alternativa~e}}}}\end{array}}$

$\boxed{\begin{array}{l}\rm 2)~ Qual~\acute e~o~conjunto~verdade~de\\\rm\ell og_x(3x+4)=\ell og_x(4x+2)?\\\tt a)~\sf0\\\tt b)~\sf1\\\tt c)~\sf2\\\tt d)~\sf3\\\tt e)~\sf4\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm igualdade~de~logaritmos}\\\sf satisfeitas~as~condic_{\!\!,}\tilde oes~de~exist\hat encia,dizemos\\\sf que~dois~logaritmos~s\tilde ao~iguais~se~e~somente~se~as~bases\\\sf e~os~logaritmandos~s\tilde ao~iguais.\\\underline{\rm matematicamente\!:}\\\sf \ell og_ba=\ell og_bc\Longleftrightarrow a=c\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm soluc_{\!\!,}\tilde ao\!:}\\\sf \ell og_x(3x+4)=\ell og_x(4x+2)\\\sf 4x+2=3x+4\\\sf 4x-3x=4-2\\\sf x=2\\\sf assim,~o~conjunto~verdade~\acute e~V=\{2\}\\\huge\boxed{\boxed{\boxed{\boxed{\sf\maltese~alternativa~c}}}}\end{array}}$

$\large\boxed{\begin{array}{l}\rm 3)~(UP). A~soluc_{\!\!,}\tilde ao~da~equac_{\!\!,}\tilde ao~logar\acute itmica\\\rm \ell og_{10}(x-4)=2~\acute e:\\\tt a)~\sf x=6\\\tt b)~\sf x=10\\\tt c)~\sf x=50\\\tt d)~\sf x=100\\\tt e)~\sf x=104\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm soluc_{\!\!,}\tilde ao\!:}\\\sf \ell og_{10}(x-4)=2\\\sf x-4=10^2\\\sf x-4=100\\\sf x=100+4\\\sf x=104\\\huge\boxed{\boxed{\boxed{\boxed{\sf\maltese~alternativa~e}}}}\end{array}}$

$\large\boxed{\begin{array}{l}\rm 4)~Resolva~a~equac_{\!\!,}\tilde ao~logar\acute itmica~\ell og_{x+3}(5x-1)=1\\\tt a)~\sf 0\\\tt b)~\sf1\\\tt c)~\sf2\\\tt d)~\sf3\\\tt e)~\sf4\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm soluc_{\!\!,}\tilde ao\!:}\\\sf admitindo~que~x>\dfrac{1}{5},vamos~resolver~a~equac_{\!\!,}\tilde ao:\\\sf \ell og_{x+3}(5x-1)=1\\\sf 5x-1=x+3\\\sf 5x-x=3+1\\\sf 4x=4\\\sf x=\dfrac{4}{4}\\\sf x=1\\\huge\boxed{\boxed{\boxed{\boxed{\sf\maltese~alternativa~b}}}}\end{array}}$