144th Problem 2017

Solution 1.

$(\frac{1}{4})^{3-2x}=8^{-x}$

$(\frac{1}{2^2})^{3-2x}=2^{3(-x})$

$2^{(-2)(3-2x)}=2^{-3x}$

$-2(3-2x)=-3x$

$-6+4x=-3x$

$7x=6$

$x=\dfrac{6}{7}$ $\large\boxed{\color{#D61F06}answer}$

Solution 2.

Take the natural logarithm of both sides.

$\ln~(\frac{1}{4})^{3-2x}=\ln~8^{-x}$

$(3-2x)~\ln\dfrac{1}{4}=-x~\ln8$

$3-2x=\dfrac{-x~\ln8}{\ln~\frac{1}{4}}$

$3-2x=-1.5x$

$3=3.5x$

$x=\dfrac{6}{7}$ $\large\boxed{\color{#D61F06}answer}$

$\displaystyle \bigg(\frac14\bigg)^{3 \ - \ 2x} = 8^{- \ x}$

$\displaystyle (2^{- 2})^{3 \ - \ 2x} = (2^3)^{- \ x}$

$\displaystyle - 2(3 - 2x) = - 3x$

$\displaystyle 4x - 6 = - 3x$

$\displaystyle 7x = 6$

$\displaystyle x = \frac 67$