Wasn't specified at the beginning that only real solutions are sought. So, lets consider the complex roots of y, w/ i=√(-1) & "e" being Euler's number. y²+4y+12=0 ⇒ y=2√3(-√(⅓)±i√(⅔))=(2√3)e^(±iθ) where cosθ=-√(⅓) & sinθ=√(⅔) ⇒ θ≅0.7π rad. x=log₂y=1+(½)log₂3±iθ/ln(2) where "ln" is the natural log.
First, learn to work with powers. 4=2^2 and 8=2^3; 4^x=(2^2)^x=(2^x)^2 and 8^x=(2^3)^x=(2^x)^3 Substitute y=2^x and rearrange to y^3+y^2-36=0. By inspection y=3, so divide by (y-3): y^2+4y+12=0 This equation has no real roots, so will consider only the case 2^x=3. Thus, x=ln3/ln2=1.58496...
U can use bit faster way also
2^3x+2^2x = 2^x(2^3+2^2) =2^x(8+4)
=2^x(12) = 36
=2^x=3
Then u can take log both side
Wasn't specified at the beginning that only real solutions are sought. So, lets consider the complex roots of y, w/ i=√(-1) & "e" being Euler's number.
y²+4y+12=0 ⇒ y=2√3(-√(⅓)±i√(⅔))=(2√3)e^(±iθ)
where cosθ=-√(⅓) & sinθ=√(⅔) ⇒ θ≅0.7π rad.
x=log₂y=1+(½)log₂3±iθ/ln(2)
where "ln" is the natural log.
8^(Log[2,1.5]+1)+4^(Log[2,1.5]+1)=36 x=Log[2,3]=Log[2,1.5]+1 final answer
First, learn to work with powers. 4=2^2 and 8=2^3; 4^x=(2^2)^x=(2^x)^2 and 8^x=(2^3)^x=(2^x)^3
Substitute y=2^x and rearrange to y^3+y^2-36=0. By inspection y=3, so divide by (y-3): y^2+4y+12=0
This equation has no real roots, so will consider only the case 2^x=3. Thus, x=ln3/ln2=1.58496...
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最初に解が実数だとは書かれていない。(虚数解もあり得る)