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Unformatted text preview: is an equation that we can easily solve. 2 x = 21 Þ x= 21
2 Now, just as with the first set of examples we need to plug this back into the original equation
and see if it will produce negative numbers or zeroes in the logarithms. If it does it can’t be a
solution and if it doesn’t then it is a solution. æ æ 21 ö ö
log 5 ç 2 ç ÷ + 4 ÷ = 2
è è2ø ø
log 5 ( 25) = 2
Only positive numbers in the logarithm and so x = 21
is in fact a solution.
[Return to Problems] (b) log x = 1 - log ( x - 3)
In this case we’ve got two logarithms in the problem so we are going to have to combine them
into a single logarithm as we did in the first set of examples. Doing this for this equation gives, log x + log ( x - 3) = 1
log ( x ( x - 3 ) ) = 1 Now, that we’ve got the equation into the proper form we convert to exponential form. Recall as
well that we’re dealing with the common logarithm here and so the base is 10.
Here is the exponential form of this equation.
© 2007 Paul Dawkins 305 http://tutorial.math.lamar.edu/terms.aspx Coll...
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- Spring '12