# Math: solving exponential equations

There are two types of exponential equations:  those that don’t need logs to be solved, and those that do.  We’ll cover the ones that don’t need logs this time; next time, we’ll cover those that do.

Consider the following example:

Solve

25^(2x-3)=(1/125)^(-x+7)

Solution:

First of all, we notice that both sides are base 5 numbers: 25=52, while 1/125=5^(-3). (You may want to brush up on negative exponents; see my post here.) The fact that both sides are in the same base means we don’t need logs; otherwise, we would.

We rewrite both sides in the common base (in this case, 5):

(5^2)^(2x-3)=(5^(-3))^(-x+7)

Recall the exponent-to-an-exponent law: (xa)b=xab. For example, (x2)5=x10. We apply that law to each side:

5^(4x-6)=5^(3x-21)

Next, we invoke another exponent law: Ap=Aq→p=q. That is, if the bases are the same, and the left side equals the right side, the exponents must be equal. (Of course, there must be no other terms, nor any arithmetic operations pending, on either side.)

Therefore,

4x-6=3x-21

We subtract 3x from both sides, then add 6 to both sides, yielding

x=-15

If you have trouble believing -15 is the answer, here’s an option: Let’s take an impartial calculator and, substituting -15 for x, carefully enter the original left side:

25^(2(-15)-3)=7.37869763×10^(-47)

Let’s do the same for the right side:

(1/125)^(-(-15)+7)=7.37869763×10^(-47)

Not necessarily everyday numbers; still, they’re equal:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.