Question

6) resolva as equações :
a) log ³na base 5 + log ( x+2) na base 5 = 2

b) log x na base 10 + log x na base 10 = 2

c ) log ( x- 3) na base 5 + log (x +2 ) na base 5 = log 14 na base 5

7) Dados log 3= 0,4771 e log 5 =0,6990 ,calcule log 0,6.

8) Resolva as equações:
a) log (x+5)- log 2=log 6

b)log (x-2) + log 3 - log 5 = log 7

9) Dados log 5 na base 7 = a e log 3 na base 7= b ,calcule :

a) log 5 na base 3.

b) log 7 na base 5.

c) log 7 na base 3.

PRECISO URGENTE!!​​

Answer 1

Explicação passo-a-passo:

6)

a)

$\sf log_{5}~3+log_{5}~(x+2)=2$

$\sf log_{5}~[3\cdot(x+2)]=2$

$\sf 3\cdot(x+2)=5^2$

$\sf 3x+6=25$

$\sf 3x=25-6$

$\sf 3x=19$

$\sf \red{x=\dfrac{19}{3}}$

b)

$\sf log_{10}~x+log_{10}~x=2$

$\sf log_{10}~(x\cdot x)=2$

$\sf log_{10}~x^2=2$

$\sf x^2=10^2$

$\sf x^2=100$

$\sf x=\sqrt{100}$

$\sf \red{x=10}$

c)

$\sf log_{5}~(x-3)+log_{5}~(x+2)=log_{5}~14$

Condição de existência:

• $\sf x-3 > 0$

$\sf x > 3$

• $\sf x+2 > 0$

$\sf x > -2$

Devemos ter $\sf x > 3$

$\sf log_{5}~[(x-3)\cdot(x+2)]=log_{5}~14$

Igualando os logaritmandos:

$\sf (x-3)\cdot(x+2)=14$

$\sf x^2+2x-3x-6=14$

$\sf x^2-x-6=14$

$\sf x^2-x-6-14=0$

$\sf x^2-x-20=0$

$\sf \Delta=(-1)^2-4\cdot1\cdot(-20)$

$\sf \Delta=1+80$

$\sf \Delta=81$

$\sf x=\dfrac{-(-1)\pm\sqrt{81}}{2\cdot1}=\dfrac{1\pm9}{2}$

$\sf x'=\dfrac{1+9}{2}~\Rightarrow~x'=\dfrac{10}{2}~\Rightarrow~x'=5$

$\sf x"=\dfrac{1-9}{2}~\Rightarrow~x"=\dfrac{-8}{2}~\Rightarrow~x"=-4$ (não serve)

Logo, $\sf x=5$

7)

$\sf log~0,6=log~\dfrac{6}{10}$

$\sf log~0,6=log~\dfrac{3}{5}$

$\sf log~0,6=log~3-log~5$

$\sf log~0,6=0,4771-0,6990$

$\sf \red{log~0,6=-0,2219}$

8)

a)

$\sf log~(x+5)-log~2=log~6$

Condição de existência:

$\sf x+5 > 0$

$\sf x > -5$

Devemos ter $\sf x > -5$

$\sf log~\left(\dfrac{x+5}{2}\right)=log~6$

Igualando os logaritmandos:

$\sf \dfrac{x+5}{2}=6$

$\sf x+5=2\cdot6$

$\sf x+5=12$

$\sf x=12-5$

$\sf \red{x=7}$

b)

$\sf log~(x-2)+log~3-log~5=log~7$

Condição de existência:

$\sf x-2 > 0$

$\sf x > 2$

Devemos ter $\sf x > 2$

$\sf log ~[(x-2)\cdot3]-log~5=log~7$

$\sf log~\left[\dfrac{(x-2)\cdot3}{5}\right]=log~7$

Igualando os logaritmandos:

$\sf \dfrac{(x-2)\cdot3}{5}=7$

$\sf \dfrac{3x-6}{5}=7$

$\sf 3x-6=5\cdot7$

$\sf 3x-6=35$

$\sf 3x=35+6$

$\sf 3x=41$

$\sf \red{x=\dfrac{41}{3}}$

9)

a)

$\sf log_{3}~5=\dfrac{log_{7}~5}{log_{7}~3}$

$\sf \red{log_{3}~5=\dfrac{a}{b}}$

b)

$\sf log_{5}~7=\dfrac{log_{7}~7}{log_{7}~5}$

$\sf \red{log_{5}~7=\dfrac{1}{a}}$

c)

$\sf log_{3}~7=\dfrac{log_{7}~7}{log_{7}~3}$

$\sf \red{log_{3}~7=\dfrac{1}{b}}$