A idade de Karina em 2020 é dada pelo valor de x na equação:
2.log₃(x+1) = log₃(21+21x)
A idade de Karina será:
a)20
b)21
c)19
d)18
e)25
A resposta é a letra E porém não estou conseguindo fazer!
Soluções para a tarefa
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2.log₃(x+1) = log₃(21+21x)
log₃(x+1)² = log₃(21+21x) ==> (x+1)² = (21+21x)
(x+1)² = 21 ==> x + 1 = 21 ==> x = 21 - 1 ==> x = 20 Letra A
Owtside:
Muito obg!
Respondido por
2
Olá Owtside,
Primeiro vamos checar a condição de existência dos logaritmandos:
![\mathsf{C.E\begin{cases}\mathsf{x+1\ \textgreater \ 0~~|~~x\ \textgreater \ -1}\\\mathsf{21+21x\ \textgreater \ 0/21~~|~~x\ \textgreater \ -1}\end{cases}}\\\\\\\\\mathsf{C.E:\{x\in\mathbb{R}~|~x\ \textgreater \ -1\}}\\\\\\\\\mathsf{2\cdot\ell og_3(x+1)=\ell og_3(21+21x)\Rightarrow \ell og_3([x+1]^2)=\ell og(21+21x)}\\\\=\\\\\mathsf{(x+1)^2=21+21x\Rightarrow (x+1)^2=21\cdot(x+1)\Rightarrow \dfrac{(x+1)^2}{(x+1)}=21}\\\\=\\\\\mathsf{x+1=21\Rightarrow \boxed{\mathsf{x=20}}}](https://tex.z-dn.net/?f=%5Cmathsf%7BC.E%5Cbegin%7Bcases%7D%5Cmathsf%7Bx%2B1%5C+%5Ctextgreater+%5C+0%7E%7E%7C%7E%7Ex%5C+%5Ctextgreater+%5C+-1%7D%5C%5C%5Cmathsf%7B21%2B21x%5C+%5Ctextgreater+%5C+0%2F21%7E%7E%7C%7E%7Ex%5C+%5Ctextgreater+%5C+-1%7D%5Cend%7Bcases%7D%7D%5C%5C%5C%5C%5C%5C%5C%5C%5Cmathsf%7BC.E%3A%5C%7Bx%5Cin%5Cmathbb%7BR%7D%7E%7C%7Ex%5C+%5Ctextgreater+%5C+-1%5C%7D%7D%5C%5C%5C%5C%5C%5C%5C%5C%5Cmathsf%7B2%5Ccdot%5Cell+og_3%28x%2B1%29%3D%5Cell+og_3%2821%2B21x%29%5CRightarrow+%5Cell+og_3%28%5Bx%2B1%5D%5E2%29%3D%5Cell+og%2821%2B21x%29%7D%5C%5C%5C%5C%3D%5C%5C%5C%5C%5Cmathsf%7B%28x%2B1%29%5E2%3D21%2B21x%5CRightarrow+%28x%2B1%29%5E2%3D21%5Ccdot%28x%2B1%29%5CRightarrow+%5Cdfrac%7B%28x%2B1%29%5E2%7D%7B%28x%2B1%29%7D%3D21%7D%5C%5C%5C%5C%3D%5C%5C%5C%5C%5Cmathsf%7Bx%2B1%3D21%5CRightarrow+%5Cboxed%7B%5Cmathsf%7Bx%3D20%7D%7D%7D "\mathsf{C.E\begin{cases}\mathsf{x+1\ \textgreater \ 0~~|~~x\ \textgreater \ -1}\\\mathsf{21+21x\ \textgreater \ 0/21~~|~~x\ \textgreater \ -1}\end{cases}}\\\\\\\\\mathsf{C.E:\{x\in\mathbb{R}~|~x\ \textgreater \ -1\}}\\\\\\\\\mathsf{2\cdot\ell og_3(x+1)=\ell og_3(21+21x)\Rightarrow \ell og_3([x+1]^2)=\ell og(21+21x)}\\\\=\\\\\mathsf{(x+1)^2=21+21x\Rightarrow (x+1)^2=21\cdot(x+1)\Rightarrow \dfrac{(x+1)^2}{(x+1)}=21}\\\\=\\\\\mathsf{x+1=21\Rightarrow \boxed{\mathsf{x=20}}}")
Resposta (a) - 20 anos
Dúvidas? comente
Primeiro vamos checar a condição de existência dos logaritmandos:
Resposta (a) - 20 anos
Dúvidas? comente
Muito obg!
Nada. Bons estudos ! :^_
:^) '
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