Question

A = log2 x=2

B= log10 x =2

C = log5 x = 6

D= log7 x = 3

E= loga 243=5

F= loga 16= -2 ​

Answer 1

Após os cálculos, temos os seguintes resultados nas equações logarítmicas da questão: A = x = 4; B = x = 100; C = x = 15625; D = x = 343; E = a = 3; F = a = 1/4.

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De antemão, vamos nos relembrar da definição de logaritmo:

                                 

                                  $\Large\quad\boldsymbol{\boxed{\boxed{\begin{array}{l}\\\ell og_a\,b=c~\Leftrightarrow~a^c=b\\\\\end{array}}}}$

Isto é, sabendo que a e b ∈ ℝ*₊ com a ≠ 1, definimos logaritmo de ''b'' numa base ''a'' sendo um número real ''c'', se e somente se ''a'' elevado a ''c'' for igual a ''b''. Assim, vamos calcular os valores das incógnitas nas eq. logarítmicas abaixo:

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$\begin{array}{l}\qquad\quad\ \ \mathsf{A=}\:\:\ell og_2\,x=2\\\\\iff~~~\mathsf{A=}\:\:2^2=x\\\\\quad\!\therefore\quad~~ \: \sf A=\:\boldsymbol{\boxed{x=4}}\end{array}$

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$\begin{array}{l}\qquad\quad\ \ \mathsf{B=}\:\:\ell og_{10}\,x=2\\\\\iff~~~\mathsf{B=}\:\:10^2=x\\\\\quad\!\therefore\quad~~ \: \sf B=\:\boldsymbol{\boxed{x=100}}\end{array}$

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$\begin{array}{l}\qquad\quad\ \ \mathsf{ C=}\:\:\ell og_5\,x=6\\\\\iff~~~\mathsf{C=}\:\:5^6=x\\\\\quad\!\therefore\quad~~\:\sf C=\:\boldsymbol{\boxed{x=15625}}\end{array}$

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$\begin{array}{l}\qquad\quad\ \ \mathsf{ D=} \: \: \ell og_7\,x=3\\\\\iff~~~\mathsf{ D=}\:\:7^3=x\\\\\quad\!\therefore\quad~~\:\sf D=\:\boldsymbol{\boxed{x=343}}\end{array}$

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$\begin{array}{l}\qquad\quad\ \ \mathsf{ E=}\:\:\ell og_a\,243=5\\\\\iff~~~\mathsf{ E=}\:\:a^5=243\\\\\iff~~~\mathsf{E=}\:\:\sqrt[5]{\:a^5~}=\sqrt[5]{\:243~}\\\\\quad\!\therefore\quad~\:~\sf E=\:\boldsymbol{\boxed{a=3}}\end{array}$

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$\begin{array}{l}\qquad\quad\ \ \mathsf{ F=}\:\:\ell og_a\,16=-\,2\\\\\iff~~~\mathsf{F=}\:\:a^{-\,2}=16\\\\\iff~~~\mathsf{F=}\:\:\dfrac{~1~}{a^2}=16\\\\\iff~~~\mathsf{F=}\:\:1=a^2\cdot16\\\\\iff~~~\mathsf{ F=}\:\:a^2=\dfrac{~1~}{16}\\\\\iff~~~\mathsf{F=}\:\:\sqrt{\:a^2~}=\sqrt{\:\dfrac{~1~}{16}~}\\\\\iff~~~\mathsf{F=}\:\:|a|=\dfrac{~1~}{4}\\\\\iff~~~\mathsf{F=}\:\:a=\pm\dfrac{~1~}{4}\end{array}$

Como a base deve ser positiva, então

$\begin{array}{l}\quad\!\therefore\quad~\:~\sf F=\:\boldsymbol{\boxed{a=\dfrac{~1~}{4}}}\end{array}$

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