Need help w/ log_3(x) + log_9(x) = 3, log_2(x) * log_4(x) *
- Thread starter oded244
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Have you conisdered a change of base?
\(\displaystyle \frac{log(x)}{log(3)} + \frac{log(x)}{log(9)} = 3\)
Or maybe
\(\displaystyle \frac{log_{3}(x)}{log_{3}(3)} + \frac{log_{3}(x)}{log_{3}(9)} = 3\)
Or notice that
\(\displaystyle \frac{log_{3}(x)}{log_{3}(3)} + \frac{log_{3}(x)}{log_{3}(9)} = 3 = log_{3}(27)\)
\(\displaystyle \frac{log(x)}{log(3)} + \frac{log(x)}{log(9)} = 3\)
Or maybe
\(\displaystyle \frac{log_{3}(x)}{log_{3}(3)} + \frac{log_{3}(x)}{log_{3}(9)} = 3\)
Or notice that
\(\displaystyle \frac{log_{3}(x)}{log_{3}(3)} + \frac{log_{3}(x)}{log_{3}(9)} = 3 = log_{3}(27)\)
Hello, oded244!
Here's another approach . . .
\(\displaystyle \text{Let: }\,\log_9(x) \:=\:Q \quad\Rightarrow\quad 9^Q \:=\:x\)
\(\displaystyle \text{Take logs, base 3: }\:\log_3(9^Q) \:=\:\log_3(x) \quad\Rightarrow\quad \log_3(3^2)^Q \:=\:\log_3(x)\)
. . \(\displaystyle \log_3(3^{2Q}) \:=\:\log_3(x) \quad\Rightarrow\quad 2Q\!\cdot\!\log_3(3) \:=\:\log_3(x) \quad\Rightarrow\quad Q \:=\:\frac{1}{2}\log_3(x)\)
\(\displaystyle \text{Hence: }\;\log_9(x) \:=\:\frac{1}{2}\log_3(x)\)
\(\displaystyle \text{Substitute into {\bf[1]}: }\;\log_3(x) + \frac{1}{2}\log_3(x) \:=\:3\)
. . \(\displaystyle \frac{3}{2}\log_3(x) \:=\:3 \quad\Rightarrow\quad \log_3(x) \:=\:2 \quad\Rightarrow\quad x \:=\:3^2\)
\(\displaystyle \text{Therefore: }\;x \;=\;9\)
Here's another approach . . .
\(\displaystyle \text{Let: }\,\log_9(x) \:=\:Q \quad\Rightarrow\quad 9^Q \:=\:x\)
\(\displaystyle \text{Take logs, base 3: }\:\log_3(9^Q) \:=\:\log_3(x) \quad\Rightarrow\quad \log_3(3^2)^Q \:=\:\log_3(x)\)
. . \(\displaystyle \log_3(3^{2Q}) \:=\:\log_3(x) \quad\Rightarrow\quad 2Q\!\cdot\!\log_3(3) \:=\:\log_3(x) \quad\Rightarrow\quad Q \:=\:\frac{1}{2}\log_3(x)\)
\(\displaystyle \text{Hence: }\;\log_9(x) \:=\:\frac{1}{2}\log_3(x)\)
\(\displaystyle \text{Substitute into {\bf[1]}: }\;\log_3(x) + \frac{1}{2}\log_3(x) \:=\:3\)
. . \(\displaystyle \frac{3}{2}\log_3(x) \:=\:3 \quad\Rightarrow\quad \log_3(x) \:=\:2 \quad\Rightarrow\quad x \:=\:3^2\)
\(\displaystyle \text{Therefore: }\;x \;=\;9\)
DR._Glockman
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To add or subtract fractions, we first need to find a common denominator. Same with logarithms, we need to find a common base. I usually use e from the natural logarithm.
Ergo: \(\displaystyle \log_2(x) + \log_4(x) + \log_8(x) = 4.5\).
\(\displaystyle Let \log_2(x) = k, then x = 2^{k}, kln(2) = ln(x), k = \frac{ln(x)}{ln(2)} = \log_2(x)\). Using the same method on the other two logarithms, we get:
\(\displaystyle \frac{ln(x)}{ln(2)} + \frac{ln(x)}{2ln(2)} + \frac{ln(x)}{3ln(2)} = 4.5\) Then
\(\displaystyle \frac{6ln(x) +3ln(x) + 2ln(x)}{6ln(2)} = 4.5\) \(\displaystyle 11ln(x) = 27ln(2), ln(x) = ln2^{27/11}, x = 2^{27/11}\)
Like differential equations (those that can be solved (implicitly or explicitly)), one can always check his answer by plugging in the results obtained, in this case, \(\displaystyle x = 2^{27/11}\).
Ergo: \(\displaystyle \log_2(x) + \log_4(x) + \log_8(x) = 4.5\).
\(\displaystyle Let \log_2(x) = k, then x = 2^{k}, kln(2) = ln(x), k = \frac{ln(x)}{ln(2)} = \log_2(x)\). Using the same method on the other two logarithms, we get:
\(\displaystyle \frac{ln(x)}{ln(2)} + \frac{ln(x)}{2ln(2)} + \frac{ln(x)}{3ln(2)} = 4.5\) Then
\(\displaystyle \frac{6ln(x) +3ln(x) + 2ln(x)}{6ln(2)} = 4.5\) \(\displaystyle 11ln(x) = 27ln(2), ln(x) = ln2^{27/11}, x = 2^{27/11}\)
Like differential equations (those that can be solved (implicitly or explicitly)), one can always check his answer by plugging in the results obtained, in this case, \(\displaystyle x = 2^{27/11}\).