Solve Log Problems

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\[\beginx & = 6x - 1\ 1 & = 5x\hspace \Rightarrow \hspacex = \frac\end\] Now, we do need to worry if this solution will produce any negative numbers or zeroes in the logarithms so the next step is to plug this into the original equation and see if it does.Since we don’t in this case we have the solution, it is \(x = \frac\).Okay, in this equation we’ve got three logarithms and we can only have two.We’ve got two logarithms on one side so we’ll combine those, drop the logarithms and then solve.\[\begin\ln \left( \right) & = \ln x\ \frac & = x\ 10 & = x\left( \right)\ 10 & = 7x - \ - 7x 10 & = 0\ \left( \right)\left( \right) & = 0\hspace \Rightarrow \hspacex = 2,\,\,x = 5\end\] We’ve got two possible solutions to check here.In this case we will use the fact that, \[x = yx = y\] In other words, if we’ve got two logs in the problem, one on either side of an equal sign and both with a coefficient of one, then we can just drop the logarithms. With this equation there are only two logarithms in the equation so it’s easy to get on one either side of the equal sign.We will also need to deal with the coefficient in front of the first term.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.\(x = 6\, :\) \[\begin\log 6 \log \left( \right) & = \log \left( \right)\ \log 6 \log 5 & = \log 30\end\] No logarithms of negative numbers and no logarithms of zero so this is a solution.\(x = - 2\, :\) \[\log \left( \right) \log \left( \right) = \log \left( \right)\] We don’t need to go any farther, there is a logarithm of a negative number in the first term (the others are also negative) and that’s all we need in order to exclude this as a solution. We are not excluding \(x = - 2\) because it is negative, that’s not the problem.

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