Solve 9^2x = 27^(1 - x) with logs
- Thread starter JAsh
- Start date
Hello, JAsh!
If we must use logs . . .
\(\displaystyle \text{Take logs: }\;\log\left(9^{2x}\right) \:=\:\log\left(27^{1-x}\right) \quad\Rightarrow\quad 2x\!\cdot\!\log(9) \:=\
1-x)\!\cdot\!\log(27)\)
. . \(\displaystyle 2x\!\cdot\!\log(9) \:=\:\log(27) - x\!\cdot\!\log(27) \quad\Rightarrow\quad 2x\!\cdot\!\log(9) + x\!\cdot\!\log(27) \:=\:\log(27)\)
\(\displaystyle \text{Factor: }\;x\left[2\log(9) + \log(27)\right] \:=\:\log(27) \quad\Rightarrow\quad x \;=\;\frac{\log(27)}{2\!\cdot\!\log(9) + \log(27)}\)
\(\displaystyle \text{Crank it through your calculator: }\;x \;=\;0.428571428571\hdots \;=\;\boxed{\frac{3}{7}}\)
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What pka was telling you: we don't need logs for this problem.
. . \(\displaystyle 9^{2x} \;=\;\left(3^{2x}\right)^2 \;=\;3^{4x}\)
. . \(\displaystyle 27^{1-x} \;=\;\left(3^3}\right)^{1-x} \;=\;3^{3(1-x)} \;=\;3^{3-3x}\)
\(\displaystyle \text{The equation becomes: }\;3^{4x} \:=\:3^{3-3x}\)
Since the bases are equal, we can equate the exponents.
. . \(\displaystyle 4x \:=\:3-3x\quad\Rightarrow\quad 7x \:=\:3\quad\Rightarrow\quad x \:=\:\frac{3}{7}\)
If we must use logs . . .
\(\displaystyle \text{Take logs: }\;\log\left(9^{2x}\right) \:=\:\log\left(27^{1-x}\right) \quad\Rightarrow\quad 2x\!\cdot\!\log(9) \:=\
1-x)\!\cdot\!\log(27)\). . \(\displaystyle 2x\!\cdot\!\log(9) \:=\:\log(27) - x\!\cdot\!\log(27) \quad\Rightarrow\quad 2x\!\cdot\!\log(9) + x\!\cdot\!\log(27) \:=\:\log(27)\)
\(\displaystyle \text{Factor: }\;x\left[2\log(9) + \log(27)\right] \:=\:\log(27) \quad\Rightarrow\quad x \;=\;\frac{\log(27)}{2\!\cdot\!\log(9) + \log(27)}\)
\(\displaystyle \text{Crank it through your calculator: }\;x \;=\;0.428571428571\hdots \;=\;\boxed{\frac{3}{7}}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
What pka was telling you: we don't need logs for this problem.
. . \(\displaystyle 9^{2x} \;=\;\left(3^{2x}\right)^2 \;=\;3^{4x}\)
. . \(\displaystyle 27^{1-x} \;=\;\left(3^3}\right)^{1-x} \;=\;3^{3(1-x)} \;=\;3^{3-3x}\)
\(\displaystyle \text{The equation becomes: }\;3^{4x} \:=\:3^{3-3x}\)
Since the bases are equal, we can equate the exponents.
. . \(\displaystyle 4x \:=\:3-3x\quad\Rightarrow\quad 7x \:=\:3\quad\Rightarrow\quad x \:=\:\frac{3}{7}\)