UPDATE: Thanks for the help but since I didn't get the confirmation in the time needed I went for tutoring instead.

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- May 8th 2008, 12:47 AMrikerConfirm My Answers? (Varied Problems)
UPDATE: Thanks for the help but since I didn't get the confirmation in the time needed I went for tutoring instead.

- May 8th 2008, 08:23 AMMathnasium
I'm not doing the whole thing, but looked at the first couple.

1. Solve 4^(x+1)=8^x

Answer: 2

Note that your answer, if we plug it back in, gives $\displaystyle 4^3 = 8^2$ which gives 64 = 64. So it's right. That's the nice thing about solving equations - you can plug it back in and see if it works. If it does, it's right!

2. Solve log(lower 2)(x-1) + log(lower 2)(x) = 3

Answer: x^2 - x - 8 = 0

What you've done so far is correct, but I assume the instructions are to solve for x, i.e., you should end up with x = some number. So you need to take what you have and solve it, probably by using the quadratic equation.

When you get your solutions, plug them back in to your original equation. If a solution results in you taking the log of a negative number, throw that solution out - you can't take the log of a negative number.

3. 7^x = 14

Answer: 1.35620718711

Looks good to me, and my calculator agrees. Also, note that, of course, 7^1 = 7 and 7^2 = 49, so your answer should be somewhere between 1 and 2 (since 14 is somewhere between 7 and 49). So it seems reasonable, too.