1. I used the intermediate value theorem for (square root x to the third)= x^2+2x-1 on (0,1), I plugged 0 and 1, and I got (left side)0=-1(right side) and 1=2 respectively. Since I got a negative(on 0), and positive on the other(on 1), there is at least one solution for (0,1), is this right? If I have the wrong idea, tell me. 2. 1/2 lnx/ln3=ln8/ln2. ans. 729. 3. (square root to the third 2)^x + 10=2^(x^2). ans. 2, -5/3 4.(ln x/ln 3)+(ln 2x+1/ln 3)=1 ans. 1 I do not know how to put the small numbers on logs so I put it in natural log form. Thanks in advance.