Germany | Can you solve this ? | A Nice Olympiad Algebra Problem

[@navay13]

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

[@kyintegralson9656]

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.

[@RyanLewis-Johnson-wq6xs]

8^(Log[2,1.5]+1)+4^(Log[2,1.5]+1)=36 x=Log[2,3]=Log[2,1.5]+1 final answer

[@wes9627]

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...

[@cachotrelles4715]

👌👌👏👏👍👏

[@gaiatetuya92]

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）