Question

log4^8+log4^4=log4^2x

2.log2(x-1)=0

log2(17+x)=5





se log 2^2=1 e log2^3=1,58 ; quanto é log18^12

Answer 1

Log₄ 8+log₄ 4=log₄ 2x
log₄ 8.4 = log₄ 2x
8.4 = 2x
x = 32/2
x = 16
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2.log₂ (x-1) = 0
log₂ (x-1)² = 0
2⁰ = (x - 1)²
1 = x² - 2x + 1
x² - 2x = 0
x . (x - 2) = 0

x' = 0
x" - 2 = 0
x" = 2
--------------------------------------------------------
log₂ (17+x) = 5
2⁵ = 17 + x
32 = 17 + x
x = 32 - 17
x = 15
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Se log₂ 2=1 e log₂ 3=1,58, quanto é log₁₈ 12 ?

$log_{18} 12 = \frac{log_{2} 12 }{log_{2}18}$
12 = 3 . 2 ²
18 = 3² . 2

$\frac{log_{2} 3.2^2 }{log_{2}3^2.2}\\ \\ \frac{log_{2} 3 + log_{2}2^2}{log_{2}3^2+log_{2}2}\\ \\ \frac{log_{2} 3 +2 log_{2}2}{2log_{2}3+log_{2}2}\\ \\ \frac{1,58 +2.(1)}{2.(1,58)+1}\\ \\ \frac{1,58 +2}{3,16+1}\\ \\ \frac{3,58}{4,16}\\ \\ 0,86$

=)