Friday, November 9, 2012
Challenging Exponents/Log Problem
Fun/challenging problem from my colleague's test for his HL students (ouch!):
2(5^(x+1)) = 1 + 3/(5^x), solve for x in exact form "a + log(b)/log(5)", where a and b are both integers.
A student asked me after school for help with this, but I couldn't quickly figure it out on the spot. Afterwards, I thought over this and found it really fun to think of a variety of ways to try approaching this problem. (Too bad they mostly didn't work.) Over dinner, I thought of one way and it worked!
I am curious how you'd solve this. Please share!
Addendum: In hindsight, it's pretty foolish of me that I never looked at the graph even though I had suggested the graphical approach to that student. It's quite a simpler problem if you consider graphical approach as aid to the algebra. I was over-complicating things by doing it completely manually. (Still possible, but takes more work, obviously.) Silly silly me. These are the kinds of automatic-habit things I'd get better at, I guess, if I started coaching ISMTF competition teams....
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That's kind of a neat problem. I'm all about the algebra, which is how I'd solve it. But I also don't like things like 5^x. I'd actually start with a substitution and make the equation look like something I'm more familiar with. So I'd make everything in terms of 5^x. So:
ReplyDelete10*5^x = 1 + 3/5^x (just extracted one of the 5s).
Substitute y = 5^x
10y = 1 + 3/y
10y^2 = y + 3 (multiply by y)
10y^2 - y - 3 = 0 (rearrange)
(5y - 3)(2y + 1) = 0 (factor)
Thus, y = 3/5 or -1/2.
Since y = 5^x, it can't be -1/2.
Thus, 5^x = 3/5
x = log(3/5)/(log(5))
Hmm, I can't imagine more than one way. I multiplied both sides by 5^x, put all terms on one side, changed my 5^(2x+1) to 5*5^2x, and then thought of it as a quadratic (y=5^x). That has a negative solution, which doesn't make sense in this problem, and a positive solution. I got 5^x=3/5, and used logs to get my final answer, which can be written as -1+log(3)/log(5). Fun problem. I may give it as a problem-solving problem to my students.
ReplyDeleteClose, Allen. Now, following the directions, how would you re-write x in the form "a + log(b)/log(5)" where a and b are both integers (no fractions)? ;)
ReplyDeleteI stop when I can plug it into my calculator! Just kidding, I am (and always have been) bad at reading directions.
ReplyDeletelog(3/5)/log(5) = [log(3) - log(5)]/log(5) = -1 + log(3)/log(5).
Interesting. I actually started with substituting "a + log(b)/log(5)" into x and working backwards using just exponent rules.
ReplyDeleteSomething like
2*5^(a + log(b)/log(5) + 1) = 1 + 3/5^(a + log(b)/log(5))
10*5^a*b = 1 + 3/(5^a*b)
10*(5^a*b)^2 = 5^a*b + 3
Substituting y for 5^a*b, we get a quadratic equation
10y^2 - y - 3 = 0
(5y - 3)(2y + 1) = 0
y = 3/5 or y = -1/2
Taking the case 5^a*b = 3/5, a = -1 and b = 3. So the solution is
x = -1 + log(3)/log(5). (I convinced myself by some simple numerical reasoning that no integer values of a and b will make 5^a*b = -1/2.)
But the graphical solution is much more elegant. If you graph both functions
f(x) = 2(5^(x+1))
g(x) = 1 + 3/(5^x), you'd see immediately that they intersect at y = 6, which means 6 = 2(5^(x+1)) and the rest of the problem becomes trivial.
Hi,
ReplyDeletein the lines of what Sue or Allen did... If you define "y" to be "5^(x + 1)" instead of "5^x", you get the answer in the form "a + log(b)/log(5)" directly (slightly more economical).
2(5^(x+1)) = 1 + 3/(5^x)
2(5^(x+1)) = 1 + 15/(5^(x+1)) multiply last term by 5/5
2y = 1 + 15/y with "y = 5^(x + 1)"
2y^2 - y - 15 = 0
which yield to...
y = -5/2 or y = 3 keeping the only y>0
5^(x + 1) = 3
log(5^(x + 1)) = log(3)
(x + 1)(log(5)) = log(3)
x + 1 = log(3)/log(5)
x = -1 + log(3)/log(5)