- #1
Elena1
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\(\displaystyle \log_{2}\left({24}\right) / \log_{96}\left({2}\right) - \log_{2}\left({192}\right) /\log_{12}\left({2}\right)\)
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Elena said:\(\displaystyle \log_{2}\left({24}\right) / \log_{96}\left({2}\right) - \log_{2}\left({192}\right) /\log_{12}\left({2}\right)\)
Elena said:i know this...in the end of book i have the solution 3 but i obtained 0 the result
Elena said:\(\displaystyle \log_{2}\left({6*{2}^{2}}\right) /\log_{6*{2}^{4}}\left({2}\right) -\log_{2}\left({3*{2}^{6}}\right)/ \log_{3*{2}^{2}}\left({2}\right)\)
Elena said:finally i obtained \(\displaystyle 4 \log_{2}\left({6}\right) / \frac{1}{8}\log_{6}\left({2}\right) -6\log_{2}\left({6}\right) /
\frac{1}{2}\log_{6}\left({2}\right)\)
Elena said:\log_{6{2}^{4}}\left({2}\right)=\frac{1}{4}\log_{3*{2}^{2}}\left({2}\right)=\frac{1}{8}\log_{6}\left({2}\right)
Elena said:\log_{6{2}^{4}}\left({2}\right)=\frac{1}{4}\log_{3*{2}^{2}}\left({2}\right)=\frac{1}{8}\log_{6}\left({2}\right)how can i write to be clear?Code:
Elena said:\(\displaystyle \log_{6*{2}^{4}}\left({2}\right)=\frac{1}{4}\log_{3*{2}^{2}}\left({2}\right)=\frac{1}{8}\log_{6}\left({2}\right)\)
how can i write to be clear?
A logarithm is the inverse function of an exponential function. It is used to solve exponential equations and is written as logb(x) where b is the base and x is the argument or input of the logarithm.
Simplifying logarithms makes them easier to work with and helps to solve complex equations. It also helps to identify patterns and relationships between numbers.
The main rules for simplifying logarithms are the product rule, quotient rule, and power rule. The product rule states that logb(xy) = logb(x) + logb(y). The quotient rule states that logb(x/y) = logb(x) - logb(y). And the power rule states that logb(xn) = nlogb(x).
Some common mistakes when simplifying logarithms include forgetting to apply the rules correctly, using the wrong base, and not simplifying the final answer. It is important to double-check your work and simplify as much as possible.
To simplify logarithms with different bases, we can use the change of base formula: logb(x) = logc(x)/logc(b), where c is any base. This allows us to convert the logarithm to a common base and then apply the rules for simplifying.