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Unformatted text preview: in the logarithms. Since we
don’t in this case we have the solution, it is x = 1
[Return to Problems] (b) log x + log ( x - 1) = log ( 3 x + 12 )
Okay, in this equation we’ve got three logarithms and we can only have two. So, we saw how to
do this kind of work in a set of examples in the previous section so we just need to do the same
thing here. It doesn’t really matter how we do this, but since one side already has one logarithm
on it we might as well combine the logs on the other side. log ( x ( x - 1) ) = log ( 3x + 12 )
Now we’ve got one logarithm on either side of the equal sign, they are the same base and have
coefficients of one so we can drop the logarithms and solve. x ( x - 1) = 3 x + 12 x 2 - x - 3x - 12 = 0
x 2 - 4 x - 12 = 0 ( x - 6 )( x + 2 ) = 0 Þ x = -2, x = 6 Now, before we declare these to be solutions we MUST check them in the original equation. x = 6: log 6 + log ( 6 - 1) = log ( 3 ( 6 ) + 12 )
log 6 + log 5 = log 30 No logarithms of negative numbers and no logarithm...
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- Spring '12