Math: solving exponential equations with logs

Tutoring math, your grade 12 students ask about logs.  The math tutor continues the discussion from last post, showing how to use logs to solve an exponential equation.

In my previous post, I showed how you might solve an exponential equation without logs when the sides can be brought to a common base.

Common base or not, you can always use logs to solve an exponential equation (provided it has a defined answer). Consider the following example:

Solve

13^(x+3)=7^(2.1x+5)

To tackle this equation, the rule of logs that you need to know is

loga^b=bloga

In other words, after you take the log, you can “bring down” the exponent to the front, where it becomes a multiple.

We take the log of each side of our equation:

log13^(x+3)=log7^(2.1x+5)

We invoke the “exponent-to-multiple” rule for logs:

(x+3)log13=(2.1x+5)log7

Apply the distributive law to each side:

xlog13+3log13=2.1xlog7+5log7

Now, we use algebra to get all the x terms to one side, then all the constants to the other. Specifically, we subtract 2.1xlog7 from both sides, then subtract 3log13 from both sides, to get

xlog13-2.1xlog7=5log7-3log13

Next, we factor out x from the left side:

x(log13-2.1log7)=5log7-3log13

Finally, we divide both sides by (log13 – 2.1log7) to isolate x on the left. We arrive at the solution

x=(5log7-3log13)/(log13-2.1log7)

If you carefully key the solution into a calculator, you’ll hopefully get -1.337334 (rounded). In order to ascertain that it truly is the solution, we substitute it for x in the original equation as follows:

13^(-1.337334+3)=7^(2.1(-1.337334)+5)

You’ll get, rounded to four decimal places,

71.1403=71.1403

Since the left side equals the right side, our solution is correct.

The logs method works whether or not the sides have a common base. Some people prefer it for every exponential equation they face.

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

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