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Solving Logarithmic Equations:
Solving with Exponentials
(page 2 of 3)

Sections: Solving from the definition, Solving with exponentials, Calculator Considerations

The second type of log equation requires the use of The Relationship:

The Relationship
 y = bx ..............is equivalent to............... (means the exact same thing as) logb(y) = x

Note that the base in both the exponential form of the equation and the logarithmic form of the equation (above) is "b", but that the x and y switch sides when you switch between the two equations. If you can remember this  that whatever had been the argument of the log becomes the "equals" and whatever had been the "equals" becomes the exponent in the exponential, and vice versa  then you should not have too much trouble with solving log equations.

• Solve log2(x) = 4.

Since this is "log equals a number", rather than "log equals log", I can solve by using The Relationship:

log2(x) = 4
24 = x

16 = x

• Solve log2(8) = x.

I can solve this by converting the logarithmic statement into its equivalent exponential form, using The Relationship:

log2(8) = x
2 x = 8

But 8 = 23, so:

2 x = 23
x = 3

Note that this could also have been solved by working directly from the definition of a logarithm: What power, when put on "2", would give you an 8? The power 3, of course!

If you wanted to give yourself a lot of work, you could also do this one in your calculator, using the change-of-base formula:

log2(8) = ln(8) / ln(2)

Plug this into your calculator, and you'll get "3" as your answer. While this change-of-base technique is not particularly useful in this case, you can see that it does work. (Try it on your calculator, if you haven't already, so you're sure you know which keys to punch, and in which order.) You will need this technique in later problems.

• Solve log2(x) + log2(x  2) = 3

I can't do anything yet, because I don't yet have "log equals a number". So I'll need to use log rules to combine the two terms on the left-hand side of the equation:

log2(x) + log2(x  2) = 3
log2((x)(x  2)) = 3

log2(x2  2x) = 3

Then I'll use The Relationship to convert the log form to the corresponding exponential form, and then I'll solve the result:

log2(x2  2x) = 3
23 = x2  2x

8 = x2  2x

0 = x2  2x  8

0 = (x  4)(x + 2)

x = 4, 2

But if x = 2, then "log2(x)", from the original logarithmic equation, will have a negative number for its argument (as will the term "log2(x  2)"). Since logs cannot have zero or negative arguments, then the solution to the original equation cannot be x = 2.

The solution is x = 4.

Keep in mind that you can check your answers to any "solving" exercise by plugging those answers back into the original equation and checking that the solution "works":

log2(x) + log2(x  2) = 3
log2(4) + log2(4  2) ?=? 3

log2(4) + log2(2) ?=? 3

Since the power that turns "2" into "4" is 2 and the power that turns "2" into "2" is "1", then we have:

log2(4) + log2(2) ?=? 3
log2(2
2) + log2(21) ?=? 3
2 + 1 ?=? 3

3 = 3

• Solve log2(log2(x))   = 1.

This may look overly-complicated, but it's just another log equation. To solve this, I'll need to apply The Relationship twice:

log2(log2(x)) = 1
21 = log2(x)
2 = log2(x)

x = 22

x = 4

Then the solution is x = 4.

• Solve log2(x2)  = (log2(x))2.

First, I'll write out the square on the right-hand side:

log2(x2) = (log2(x))2
log2(x2) = (log2(x)) (log2(x))

Then I'll apply the log rule to move the "squared", from inside the log on the left-hand side of the equation, out in front of that log as a multiplier. Then I'll move that term to the right-hand side:

2log2(x) = [log2(x)] [log2(x)]
0 = [log2(x)] [log2(x)]    2log2(x)

This may look bad, but it's nothing more than a factoring exercise at this point. So I'll factor, and then I'll solve the factors by using The Relationship:

0 = [log2(x)] [log2(x)  2]
log2(x) = 0  or  log2(x)  2 = 0

20 = x   or  log2(x) = 2

1 = x  or  22 = x

1 = x  or  4 = x

The solution is x = 1, 4.

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 Cite this article as: Stapel, Elizabeth. "Solving Logarithmic Equations with Exponentials." Purplemath. Available from     http://www.purplemath.com/modules/solvelog2.htm. Accessed [Date] [Month] 2016

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