Russian l can you solve this exponential problem?? l Olympiad Math

[@1234larry1]

It might be better if instead of writing every step like 2*1=2, you took that time to verify at least one of the complex solutions.

[@Luiz30072]

I thought so too because, Larry, making u-2=0, u=2. Now just check u=7^x=2 and we will get the right answer. So, why use log?

[@prollysine]

let u=7^x , u^3+u^2+u-14=0 , (u-2)(u^2+3u+7)=0 , u=2 , 7^x=2 , x=log2/log7 , test , 7^(log2/log7)+7^(2*log2/log7)+7^(3*log2/log7)=2+4+8 , --> 14 , OK ,

[@ToanPham-wr7xe]

😮

[@ConradoPeter-hl5ij]

a=7ⁿ et a > 0 (I) a+a²+a³=14 a³+a²+a-14=0 (a³-a)-15+(a²+2a+1)=0 a(a²-1)+(a+1)²=15 (a+1)[a(a-1)+(a+1)]=15 (a+1)[a²-a+a+1]=3*5 (a+1)[a²+1]=(2+1)(2²+1) a=2. (II) Briot-Ruffinni 1 1 1 -14 | 2 1 3 7 0 (a-2)[a²+3a+7]=0 a²+3a+7=0 a=-(3/2)±i[(19½)/2] (III et IV) Sed a > 0. Ergo: I et II, Ergo: 7ⁿ=2 n*log7=log2 n=[log2]/[log7] n ≈ 0,30/ 0,84 n ≈ 30/84 n ≈ 0,356