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....


  1. 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:

    10*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))

  2. 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.

  3. Close, 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)? ;)

  4. I stop when I can plug it into my calculator! Just kidding, I am (and always have been) bad at reading directions.

    log(3/5)/log(5) = [log(3) - log(5)]/log(5) = -1 + log(3)/log(5).

  5. Interesting. I actually started with substituting "a + log(b)/log(5)" into x and working backwards using just exponent rules.

    Something 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.

  6. Hi,

    in 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)