como resolver equacoes exponencial com raiz?
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Boa noite Elias!
Solução!
![9^{(x+1)}= \sqrt[3]{3}\\\\\\
3^{2} ^{(x+1)}= (3)^{ \frac{1}{3} } \\\\\\
3 ^{(2x+2)}= (3)^{ \frac{1}{3} } \\\\\\
2x+2= \frac{1}{3}\\\\\
6x+6=1\\\\\
6x=1-6\\\\
6x=-5\\\\\\
x= -\frac{5}{6}\\\\\
x=- \frac{5}{6}](https://tex.z-dn.net/?f=9%5E%7B%28x%2B1%29%7D%3D+%5Csqrt%5B3%5D%7B3%7D%5C%5C%5C%5C%5C%5C++%0A3%5E%7B2%7D+%5E%7B%28x%2B1%29%7D%3D+%283%29%5E%7B+%5Cfrac%7B1%7D%7B3%7D+%7D+%5C%5C%5C%5C%5C%5C%0A++3+%5E%7B%282x%2B2%29%7D%3D+%283%29%5E%7B+%5Cfrac%7B1%7D%7B3%7D+%7D+%5C%5C%5C%5C%5C%5C%0A2x%2B2%3D+%5Cfrac%7B1%7D%7B3%7D%5C%5C%5C%5C%5C%0A6x%2B6%3D1%5C%5C%5C%5C%5C%0A6x%3D1-6%5C%5C%5C%5C%0A6x%3D-5%5C%5C%5C%5C%5C%5C%0Ax%3D+-%5Cfrac%7B5%7D%7B6%7D%5C%5C%5C%5C%5C%0Ax%3D-+%5Cfrac%7B5%7D%7B6%7D+++++ "9^{(x+1)}= \sqrt[3]{3}\\\\\\
3^{2} ^{(x+1)}= (3)^{ \frac{1}{3} } \\\\\\
3 ^{(2x+2)}= (3)^{ \frac{1}{3} } \\\\\\
2x+2= \frac{1}{3}\\\\\
6x+6=1\\\\\
6x=1-6\\\\
6x=-5\\\\\\
x= -\frac{5}{6}\\\\\
x=- \frac{5}{6}")
Boa noite!
Bons estudos!
Solução!
Boa noite!
Bons estudos!
Respondido por
1
9ˣ⁺¹= ∛3
9ˣ⁺¹= 3¹/³
(3²)ˣ⁺¹= 3¹/³
2x+2=1/3
3×(2x+2) = 1
6x+6 = 1
6x =1-6
6x= -5
x = -5/6
9ˣ⁺¹= 3¹/³
(3²)ˣ⁺¹= 3¹/³
2x+2=1/3
3×(2x+2) = 1
6x+6 = 1
6x =1-6
6x= -5
x = -5/6
Saccaniana:
Não tenho certeza!
eu conferi ta certo sim!
obrigado, entendi já, mais a minha questão tem 3 em cima da raiz , então e só fazer o msm procedimento , so q trocando o 2 por 3!
acho q e assim!
sim, vc tem que igualar as bases.
valew.
valew
valew.
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