Solve 9^2x = 27^(1 - x) with logs

[JAsh]

solve with logs: 9^2x = 27^(1-x)

please help!

 

[pka]

Re: 9^2x = 27^(1-x)

$\displaystyle 3^{4x} = 9^{2x} = 27^{\left( {1 - x} \right)} = 3^{\left( {3 - 3x} \right)}$

 

[soroban]

Hello, JAsh!

Solve with logs: .$\displaystyle 9^{2x} \:= \:27^{1-x}$

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.

. . $\displaystyle 4x \:=\:3-3x\quad\Rightarrow\quad 7x \:=\:3\quad\Rightarrow\quad x \:=\:\frac{3}{7}$