Solving for x using logarithms

1. Jun 12, 2012

Opus_723

1. The problem statement, all variables and given/known data

I tutor math for a couple of high school kids, and usually don't have any problems. Occasionally we run into a problem that takes me a minute, since I haven't actually used a lot of this stuff since I was in high school, but I always figure it out very quickly when that happens.

However, yesterday I came across a problem in the middle of a set of really straightforward logarithm problems, and I couldn't work out how to do it. It was a little embarassing, but I'm more curious than anything. How would one go about solving an equation like this?

30(1.4)$^{x}$ = 30 + 0.4x

Of course, the numbers aren't important, just the general layout, with an x in an exponent and in a term.

No matter how I work it, I can't seem to isolate the x. It seems like there must be some simple approach, especially since all of the other problems were so easy, but I can't think of it.

2. Jun 12, 2012

dalcde

3. Jun 12, 2012

clamtrox

You have to be kind of clever to solve this. Here are possible steps:

1) show there are two solutions
2) show that x=0 is a solution
3) second solution is roughly where 30+0.4x is "small" -> x ~-75 -> 1.4^x ~ 0 so second solution is x≈-75.

In general, you can't solve this kind of equations analytically.

4. Jun 12, 2012

Opus_723

How can you show that there are two solutions, and that one is where x is "small"?

5. Jun 12, 2012

vela

Staff Emeritus
Graph the two sides of the equation. You can quickly see there will be two solutions and that one of them is near the x-axis.

6. Jun 13, 2012

Ratch

Opus 723,

There are many equations and are not amenable to a closed form solution. For instance, cos(x) = x. You can use approximation methods like the Newton-Raphson mentioded earlier, or you can express a term in the expression with a few terms of a fast converging series, and solve the equation that way. Cos(x) in the example I gave can be expressed as a series.

Ratch

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