Question

Como resolver essas equações exponenciais?

Answer 1

a)
$10^{1-4x}=0,001\\ \\ 10^{1-4x}=10^{-3}\\ \\ 1-4x=-3\\ \\ -4x=-4\\ \\ x=1$

b)
$6*7^{x+2}=294\\ \\ 7^{x+2}=\frac{294}{6}\\ \\ 7^{x+2}=49\\ \\ 7^{x+2}=7^2\\ \\ x+2=2\\ \\ x=0$

c)
$5^{2-x}=\frac{1}{125}\\ \\ 5^{2-x}=\frac{1}{5^3}\\ \\ 5^{2-x}=5^{-3}\\ \\ 2-x=-3\\ \\ -x=-3-2\\ \\ -x=-5\\ \\ x=5$

d)
$8^{2-x}=(0,25)^{x-1}\\ \\ 8^{2-x}=(\frac{1}{4})^{x-1}\\ \\ (2^3)^{2-x}=(\frac{1}{2^2})^{x-1}\\ \\ 2^{6-3x}=(2^{-2})^{x-1}\\ \\ 2^{6-3x}=2^{-2x+2}\\ \\ 6-3x=-2x+2\\ \\ x=4$

Answer 2

$a) 10^{1-4x} = 10^{-3} \to 1-4x=-3\to 1+3=4x\to 4x=4 \therefore x=1 \\ \\b) 7^{x+2}= \dfrac{294}{6} \to 7^{x+2} =49\to 7^{x+2}= 7^{2} \to x+2=2 \therefore x=0 \\ \\ c) 5^{2-x}= \dfrac{1}{ 5^{3} } \to 5^{2-x}= 5^{-3}\to 2-x=-3\to 2+3=x ~~~\therefore ~~x=5 \\ \\d) (2^{3})^{2-x} = (\dfrac{1}{4}) ^{x+1}\to 2^{6-3x} =( \dfrac{1}{2^{2} } )^{x+1}\to 2^{6-3x}= (2^{-2} )^{x+1}\to \\ \\ 6-3x=-2x-2\to 6+2=3x-2x~~\therefore ~~~~x=8 \\ \\.$