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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:

Note that the base in both the exponential equation and the log equation (above) is "b", but that the x and y switch sides when you switch between the two equations. If you can remember this (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 terms on the left-hand side:

log₂(x) + log₂(x – 2) = 3
log₂((x)(x – 2)) = 3
log₂(x² – 2x) = 3

Then use The Relationship:

log₂(x² – 2x) = 3
2³ = x² – 2x
8 = x² – 2x
0 = x² – 2x – 8
0 = (x – 4)(x + 2)
x = 4, –2

Since "log₂(x)", from the original logarithmic equation, cannot have a negative number for its argument (and since "log₂(x – 2)" cannot have zero for its argument), then the solution cannot be x = –2.

The solution is x = 4.

Always remember that, whenever you solve any type of equation, you can always check your answers by plugging them 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
2 + 1 ?=? 3
3 = 3

The solution checks. Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

• Solve log₂(log₂(x))   = 1.

Don't panic! Solving this is going to require using 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.

Write out the square on the right-hand side:

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

Apply the log rule to move the square outside as a multiplier, and move the term to the right-hand side:

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

Factor, and solve, 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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