Let a=8^x. So, a is positive. The equation is written as (a^2-9)^1/3 = a/2 > a^3-8a^2+72=0 > (a-6)(a^2-2a-12)=0. So, a=6 and a=√(13) +1. Both are valid solutions. Thus, x=ln6/ln8= 1/(3ln2)[ln 3 + ln 2] and x = 1/(3ln2)[ln(√13 +1)].
Para a maioria dos brasileiros é muito difícil ter essa percepção, em fazer -8y² = -6y² - 2y². (For most Brazilians, it is very difficult to have this perception, to do -8y² = -6y² - 2y².)
Η εξισωση γραφεται [(8^χ)^2-3^2]^(1/3)=[([(8^χ)^2]^(1/2)]/2. Αν 8^χ= y >0 εχω y^3-8y^2 +72=0 ή (y-6)(y^2-2y-12)=0....y=6 y=1+(13)^1/2 y>0 τελικα χ=log6/log8=[ log(2×3)]/3log2=(log2+ log3)/3log2 ή χ=[log (1+ριζα13)]/log8
Let a=8^x. So, a is positive. The equation is written as (a^2-9)^1/3 = a/2 > a^3-8a^2+72=0 > (a-6)(a^2-2a-12)=0. So, a=6 and a=√(13) +1. Both are valid solutions. Thus, x=ln6/ln8= 1/(3ln2)[ln 3 + ln 2] and x = 1/(3ln2)[ln(√13 +1)].
X= 1/3{(log3/log2)+1};{log(√13+1)}/ log8
Para a maioria dos brasileiros é muito difícil ter essa percepção, em fazer -8y² = -6y² - 2y². (For most Brazilians, it is very difficult to have this perception, to do -8y² = -6y² - 2y².)
Η εξισωση γραφεται [(8^χ)^2-3^2]^(1/3)=[([(8^χ)^2]^(1/2)]/2. Αν 8^χ= y >0 εχω y^3-8y^2 +72=0 ή (y-6)(y^2-2y-12)=0....y=6 y=1+(13)^1/2 y>0 τελικα χ=log6/log8=[ log(2×3)]/3log2=(log2+ log3)/3log2 ή χ=[log (1+ριζα13)]/log8
[(8^x + 3)(8^x - 3)]1/3 = √(4^(3x-1) = y
=> (8^x + 3)(8^x - 3) = y^3 (*)
and 4^(3x-1) = y^2 (**) .
From (*) => (8^x)^2 - 3^2 = y^3 =>
64^x - 9 = y^3 => (4^3)^x - 9 = y^3 =>
4^(3x) = 9 + y^3 (*)' .
From (**) y^2 = (4^(3x))/4 => 4^(3x) = 4• y^2 (**)' .
From (*)' and (**)' => 9 + y^3 = 4 y^2 => y^3 - 4y^2 + 9 = 0 =>
(y - 3)(y^2 - y + 3) = 0 =>
y = 3 and y = (- 1 ±√13).
Thus 4^(3x-1) = y = 3 => 4^(3x) = 12
=> log (4^(3x)) = log 12 =>
(3x) log 4 = log 12 => x = 1/3 •[(log 12)/(log4)] = ..
For y = ( -1 +√13)/2 => ..
x = 1/3 • [(log ( - 1 +√13))/(log 4)] .
For y = - 1 - √13 < 0 the solution doesn't is exist because 4^(3x -1) > 0