Logarithm Q: 2logx - 1/2log(x-1) + 3log(x+2) = 0
- Thread starter Spoon-
- Start date
Re: Logarithm Question
Just to add, reviewing these may help:
\(\displaystyle nlog_{a}b = log_{a}b^{n}\)
\(\displaystyle a^{\frac{m}{n}} = \sqrt[m]{a^{n}} = (\sqrt[m]{a})^{n}\)
\(\displaystyle log_{a}b + log_{a}c = log_{a}(b \cdot c)\)
\(\displaystyle log_{a}b - log_{a}c = log_{a}\left(\frac{b}{c}\right)\)
\(\displaystyle log_{a}b = c \quad \mbox{is the same thing as} \quad a^{c} = b\)
Just to add, reviewing these may help:
\(\displaystyle nlog_{a}b = log_{a}b^{n}\)
\(\displaystyle a^{\frac{m}{n}} = \sqrt[m]{a^{n}} = (\sqrt[m]{a})^{n}\)
\(\displaystyle log_{a}b + log_{a}c = log_{a}(b \cdot c)\)
\(\displaystyle log_{a}b - log_{a}c = log_{a}\left(\frac{b}{c}\right)\)
\(\displaystyle log_{a}b = c \quad \mbox{is the same thing as} \quad a^{c} = b\)
D
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Spoon- said:
Did you really want to solve for 'x'?
What methods have learned to solve non-linear (beyond quadratic) equations?