Question

Determine o valor das incógnitas a, b e c:

a) $log_{2}$ a = 2
b) $log_{25}$ $5^{b}$ = b + 1
c) c · $log_{9}$ 3 = 2c + 1

Answer 1

a) $log_{2}a = 2$

2² = a

a = 4

b)$log_{25} 5^{b} = b+1$

$b. log_{25}5 = b+1$

$b. log_{25} 25^{1/2} = b+1$

b/2 = b+1

b/2 = -1

b = -2


c) $c.log_{9}3 = 2c+1$

$c. log_{9} 9^{1/2} = 2c+1$

$c/2 = 2c+1$

3c/2 = -1

c = -2/3

Answer 2

Ae mano,

transforme as expressões logarítmicas, em exponenciais, usando a definição

$\log_x(y)=n~~\Rightarrow ~~x^n=y$

E a 3a propriedade de logaritmo, a da potência:

$n\cdot\log_x(y)=\log_x(y)^n$

facin!

$\log_2(a)=2\\\\ 2^2=a\\\\ \huge\boxed{a=4}$

.........................

$\log_{25}(5^b)=b+1\\\\ 25^{b+1}=5^b\\ (5^2)^{b+1}=5^b\\ 5^{2b+2}=5^b\\ \not5^{2b+2}=\not5^b\\\\ 2b+2=b\\ b-2b=2\\ -b=2~~(multiplique~por~-1)\\\\ \huge\boxed{b=-2}$

.........................

$c\cdot\log_9(3)=2c+1\\ \log_9(3)^c=2c+1\\\\ 9^{2c+1}=3^c\\ (3^2)^{2c+1}=3^c\\ 3^{4c+2}=3^c\\ \not3^{4c+2}=\not3^c\\\\ 4c+2=c\\ c-4c=2\\ -3c=2\\\\ \Large\boxed{c= -\dfrac{2}{3} }$

Tenha ótimos estudos ;P