I was tutoring a pre-calculus student today, and she had a homework problem I didn't know how to do. I played it smooth, saying, "Well, if you're not sure how to do it, let's check through the book for an example." So here's the problem. Go:
solve for x by algebra
32x + 3x+1 - 4 = 0
Math
Subscribe to:
Post Comments (Atom)
3 comments:
Uh, x=0?
I took 14 hours, but that still makes me first. I guess the first half the problem is how often you check Mark's blog...
I'm a winner!
how about 3^(2x)+3^(x+1)-648=0
Wait. Is this as simple as I thought or did I make a mistake somewhere?
Since this is for a pre-algebra class or something, it can't be too tricky.
Take 3^(2x)+3^(x+1)-648=0, can be written also like:
(3^x)^2+3*(3^x)-648=0
Let y=x^2. We get:
y^2+3y-648=0 which can be factored:
(y+27)(y-24)=0
Using the Zero Product Property we get the solutions
y=-27, y-24. Plugging back 3^x, we get,
3^x=-27, 3^x=24. The first solution is garbage and we don't like dealing with that in pre-algebra so we cross it out and move onto the next one.
3^x=24 is our solution and now we solve for x.
Rewriting in exponential:
e^(xln(3))=24
Take natural log of both sides:
ln(e^(xln(3)))=ln(24)
xln(3)=ln(24)
x=(ln(24)/ln(3))
We could do the same thing with the first problem, I think, but I am currently lacking my Ti-83 Plus, so I can't even confirm if this answer is correct.
Can someone check for me?
Post a Comment