Question

determinar x nas equações exponenciais:
${25}^{x} = \sqrt{5}$

$(\frac{1}{9} ) {}^{x} = \sqrt{27}$



${49}^{2x - 3} = 343$


$( \frac{5}{3} ) {}^{x} = \frac{2}{125}$
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Answer 1

A) 25^x=V5
5^(2x)=5^(1/2)
2x=1/2
x=1/4

B) (1/9)^x=V27
9^(-x)= 3^(3/2)
3^(-2x)= 3^(3/2)
-2x=3/2
x=-3/4

C) 49^(2x-3)=343
7^(4x-6)=7^3
4x-6=3
x=9/4

D) (5/3)^x=2/125
Log5/3((5/3)^x)=log5/3(2/125)
x=log5/3(2/125)