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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 logarithmic equation has two logs, each having the same base, 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 their arguments to be equal. In other words, the log expressions being equal says that the arguments must be equal, so I have:

x = 14

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And that's 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 I can solve the log equation by solving this quadratic equation:

x² – 2x + 1 = 0
(x – 1)(x – 1) = 0

Then the solution is x = 1.

Logarithms cannot have non-positive arguments, but quadratics and other equations can have negative solutions. So it is generally a good idea to check the solutions you get for log equations:

   log_b(x²)  =   log_b(2x – 1)
log_b([1]²) ?=? log_b(2[1] – 1)
    log_b(1) ?=? log_b(2 – 1)
    log_b(1)  =   log_b(1)

The value of the base of the log is irrelevant here. Each log has the same base, each log ends up with the same argument, and that argument is a positive value, so the solution "checks".

• 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 Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

Since I cannot have a negative inside a logarithm, the quadratic-equation solution "x = –5" can not be a valid solution to the original logarithmic equation (in particular, this negative value won't work in the right-hand side of the original equation).

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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