Question

calcule os seguintes logaritmos
A) 1/9 3 √3
B) log7 √7/√49
C) 125/27 0,6
Alguem me ajuda????
​

Answer 1

Explicação passo-a-passo:

a)

$\sf log_{\frac{1}{9}}~3\sqrt{3}=x$

$\sf \Big(\dfrac{1}{9}\Big)^x=3\sqrt{3}$

$\sf \Big(\dfrac{1}{3^2}\Big)^x=\sqrt{3^2\cdot3}$

$\sf (3^{-2})^x=\sqrt{3^3}$

$\sf 3^{-2x}=3^{\frac{3}{2}}$

Igualando os expoentes:

$\sf -2x=\dfrac{3}{2}$

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

$\sf -4x=3~~~~~\cdot(-1)$

$\sf 4x=-3$

$\sf \red{x=\dfrac{-3}{4}}$

b)

$\sf log_{7}~\Big(\dfrac{\sqrt{7}}{\sqrt{49}}\Big)=x$

$\sf 7^x=\dfrac{\sqrt{7}}{\sqrt{49}}$

$\sf 7^x=\dfrac{\sqrt{7}}{7}$

$\sf 7^x=\dfrac{7^{\frac{1}{2}}}{7^1}$

$\sf 7^x=7^{\frac{1}{2}-1}$

$\sf 7^x=7^{\frac{1-2}{2}}$

$\sf 7^x=7^{\frac{-1}{2}}$

Igualando os expoentes:

$\sf \red{x=\dfrac{-1}{2}}$

c)

$\sf log_{\frac{125}{27}}~0,6=x$

$\sf \Big(\dfrac{125}{27}\Big)^x=0,6$

$\sf \Big(\dfrac{5^3}{3^3}\Big)^x=\dfrac{3}{5}$

$\sf \Big[\Big(\dfrac{5}{3}\Big)^3\Big]^x=\dfrac{3}{5}$

$\sf \Big(\dfrac{5}{3}\Big)^{3x}=\dfrac{3}{5}$

$\sf \Big(\dfrac{5}{3}\Big)^{3x}=\Big(\dfrac{5}{3}\Big)^{-1}$

Igualando os expoentes:

$\sf 3x=-1$

$\sf \red{x=\dfrac{-1}{3}}$