Question

Lg³(x+1)+lg²(x+1)-2lg(x+1)=0

Answer 1

1) log324-log38=log3(24:8)=log33=1;2) log5600-log512-log550=log5(600:12:50)=log51=0;3) lg800-lg25-lg3,2=lg (800:25:3,2)=lg10=1.Найти: 1) log252,5, если log2510=a.log252,5=log25(25:10)=log2525-log2510=1-a;2) lg0,73, если lg7,3=b.lg0,73=lg (7,3:10)=lg7,3-lg10=b-1;3) ln (e/5), если ln5=c.ln (e/5)=lne-ln5=1-c.

Answer 2

I can help you with your math problem. Here is the detailed solution:

The given equation is $$\log^3(x+1)+\log^2(x+1)-2\log(x+1)=0$$ We can rewrite it as $$\left(\log(x+1)\right)^3+\left(\log(x+1)\right)^2-2\log(x+1)=0$$ Then we can factor it as $$\log(x+1)\left(\log^2(x+1)+\log(x+1)-2\right)=0$$ Using the quadratic formula, we can find the roots of the quadratic expression as $$\log(x+1)=-2 \text{ or } \log(x+1)=1$$ Solving for x, we get $$x+1=e^{-2} \text{ or } x+1=e^1$$ Therefore, $$x=e^{-2}-1 \text{ or } x=e^1-1$$ These are the solutions of the original equation.Ответь+подробно.+Lg³(x+1)+lg²(x+1)-2lg(x+1)=0