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Solving Logarithmic Equations:
     Solving from the Definition (page 1 of 3)

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

The first type of log equation has two logs set equal to each other, and you solve by setting the insides (the "arguments") equal to each other. For example:

• Solve log₂(x) = log₂(14).

Since the logarithms on either side of the equation have the same base ("2", in this case), then the only way these two logs can be equal is for the arguments to be equal. That is, it must be that:

x = 14

And this is the solution: x = 14

• Solve log_b(x²) = log_b(2x – 1).

Since the bases of the logs are the same (the unknown value"b", in this case), then the insides must be equal. That is:

x² = 2x – 1

Then:

x² – 2x + 1 = 0
(x – 1)(x – 1) = 0

Then the solution is x = 1.

• Solve log_b(x² – 30) = log_b(x).

Since the logs have the same base, I can set the arguments equal and solve:

x² – 30 = x
x² – x – 30 = 0
(x – 6)(x + 5) = 0
x = 6, – 5

Since you cannot have a negative inside a logarithm, the solution "x = –5" is not a valid solution to the original logarithmic equation. Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

The solution is x = 6.

• Solve 2log_b(x) = log_b(4) + log_b(x – 1).

All of these logs have the same base, but I can't solve yet, because I don't yet have "log equals log". So first I'll have to apply log rules:

2log_b(x) = log_b(4) + log_b(x – 1)
log_b(x²) = log_b((4)(x – 1))
log_b(x²) = log_b(4x – 4)

Then:

x² = 4x – 4
x² – 4x + 4 = 0
(x – 2)(x – 2) = 0

The solution is x = 2.

• Solve ln( e^x ) = ln( e³ ) + ln( e⁵ ).

Remember the defintion of logarithms. Logarithms are powers. Specifically, "log_b(a)" is the power that, when put on the base "b", gives you "a". In this case, the base of the log is e. The argument of "ln( e^x )" is "e^x". That is, "ln( e^x )" is "the power that, when put on e, gives you e^x.

Well, what power do you have to put on e to get e^x? Why, x, of course! So ln( e^x ) = x. Similarly, ln( e³ ) = 3 and ln( e⁵ ) = 5. So the given equation simplifies quite nicely:

ln( e^x ) = ln( e³ ) + ln( e⁵ )
x = 3 + 5
x = 8

The solution is x = 8.

Note: This could also have been solved using log rules:

ln( e^x ) = ln( e³ ) + ln( e⁵ )
ln( e^x ) = ln(( e³ )( e⁵ ))
ln( e^x ) = ln( e^{3 + 5} )
ln( e^x ) = ln( e⁸ )

Comparing the arguments:

e^x = e⁸
x = 8

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