Question

8) Resolva as equações:
a) log(x^2)=(logx)^2
b)log2 ^3Vx + log8 X + log64 X^2 = 1/3

preciso de ajuda URGENTE!!!!!

Answer 1

$\boxed{\begin{array}{l}\tt a)~\ell og(x^2)=(\ell og x)^2\\\sf 2\ell ogx=(\ell ogx)^2\\\sf(\ell ogx)^2-2\ell ogx=0\\\sf\ell ogx(\ell ogx-2)=0\\\sf\ell ogx=0\\\sf x=10^0=1\\\sf\ell ogx-2=0\\\sf\ell ogx=2\\\sf x=10^2\\\sf x=100\\\sf S=\{1,100\}\end{array}}$

$\boxed{\begin{array}{l}\sf\ell og_2\sqrt[\sf3]{\sf x}+\ell og_8x+\ell og_{64}x^2=\dfrac{1}{3}\\\sf\ell og_2x^{\frac{1}{3}}+\ell og_{2^3}x+\ell og_{2^6}x^2=\dfrac{1}{3}\\\sf\dfrac{1}{3}\ell og_2x+\dfrac{1}{3}\ell og_2x+\dfrac{1}{6}\cdot2\ell og_2x=\dfrac{1}{3}\end{array}}$

$\boxed{\begin{array}{l}\sf\dfrac{1}{3}\ell og_2x+\dfrac{1}{3}\ell og_2x+\dfrac{1}{3}\ell og_2x=\dfrac{1}{3}\\\sf\ell og_2x=\dfrac{1}{3}\\\sf x=2^{\frac{1}{3}}\\\sf x=\sqrt[\sf3]{\sf 2}\end{array}}$

Answer 2

Resposta:

a) ㏒(x²) = (㏒x)²

2 · ㏒x = ㏒x · ㏒x , como ㏒x ≠ 0 podemos dividir dos dois lados

2 = logx ⇒ 10² = x ⇒ x = 100.

b)

b)$log(2^{3} \sqrt{x})+log(8x)+log(64x^{2} ) = 1/3\\log(2^{3}x^{\frac{1}{2} })+ log(2^{3}x )+ log(2^{6}x^{2} )= 1/3\\log(2^{3}\cdot2^{3}\cdot2^{6}\cdot x^{7/2})=1/3\\log(2^{12}\cdot x^{7/2} )=1/3\\log2^{12} + logx^{7/2} = 1/3\\12\cdot log2 + 7/2\cdot logx = 1/3\\36 \cdot log 2 + 21/2\cdot log x = 1\\21 \cdot logx = (1-36\cdot log2)\cdot2\\logx = ( 1- 36\cdot log2) 2\cdot1/21$

Explicação passo a passo:

eu acho que a armação da b) não tá correta acho que tá errado a raiz quadrada de x