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Solving Exponential Equations:
     Solving by Using Logarithms (page 2 of 3)

Sections: Solving from the definition, Solving using logarithms, Calculators

Most exponential equations do not solve neatly, in which case you need to use logarithms. For instance:

• Solve 2^x = 30.

If this had said "Solve 2^x = 32", it would have been easy, because I could have converted the 32 to 2⁵, set the exponents equal, and found my solution of "x = 5". But 30 is not a power of 2, so I can't set powers equal to each other. Then I need some other method of getting at the x, because I can't solve with it floating up there above the 2; I need it back down on the ground where it belongs. And I'll have to use logarithms to get at it.

Remember that, when dealing with equations, you can do whatever you like to the equation, as long as you do the same thing to both sides. Remember also that, to solve an equation, you have to get the variable by itself on one side of the "equals" sign, and that you isolate the variable by undoing whatever has been done to it.

In this case, the variable x has been put in the exponent. The backwards (technically, the "inverse") of exponentials are logarithms, so I'll need to undo the exponent by taking the log of both sides of the equation. This will work because of the log rule that says that exponents inside a log can be turned into multipliers in front of the log:

log_b(mⁿ) = n · log_b(m)

I can take any log I like (base-10 log, base-2 log, natural log, etc), but some are more useful than others. Since the base in the equation "2^x = 30" is "2", you might try log-base-2:

log₂(2^x) = log₂(30)
xlog₂(2) = log₂(30)
x(1) = log₂(30)
x = log₂(30)

But I can't evaluate this in my calculator unless I apply the change-of-base formula:

x = log₂(30)
   = ^ln(30)/_ln(2)

What would happen if I just used the natural log in the first place?

2^x = 30
ln(2^x) = ln(30)
xln(2) = ln(30)
x = ^ln(30)/_ln(2)

Either way, I get the same answer, but taking natural log in the first place was simpler and shorter.

Note: I could have used the common (base-10) log instead of the natural log, and still come up with the same value (when evaluated in the calculator). Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

Since science uses the natural log so much, and since it is one of the two logs that calculators can evaluate, I tend to take the natural log of both sides when solving exponential equations. This is not (generally) required, but is often more useful than other options.

• Solve 5^x = 212.

Since 212 is not a power of 5, then I will have to use logs to solve this equation. I could take base-5 log of each side, solve, and then apply the change-of-base formula, but I think I'd rather just use the natural log in the first place:

5^x = 212
ln(5^x) = ln(212)
xln(5) = ln(212)
x = ^ln(212)/_ln(5)

...or about 3.328, rounded to three decimal places.

• Solve 10^2x = 52.

Since 52 is not a power of 10, I will have to use logs to solve this.  In this particular instance, since the base is 10 and since base-10 logs can be done on the calculator, I will use the common log instead of the natural log to solve this equation:

10^2x = 52
log(10^2x) = log(52)
2xlog(10) = log(52)
2x(1) = log(52)
2x = log(52)
x = ^log(52)/₂

...or about 0.858, rounded to three decimal places.

• Solve 3(2^x+4) = 350.

Before I can start looking at the exponential, I first have to get rid of the 3, so I'll divide that off to get:

2^x+4 = ³⁵⁰/₃

Since ³⁵⁰/₃ is not a power of 2, I will have to use logs. I will use the natural log in this case:

...or about 2.866, rounded to three decimal places.

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