# How do you solve 4^(3-2x)=5^-x?

Jul 26, 2017

x=((3log4)/(2log4-log5))≈3.5755

#### Explanation:

${4}^{3 - 2 x} = {5}^{-} x$

1. Take the common logarithm of both sides
$\log {4}^{3 - 2 x} = \log {5}^{-} x$
2. Move the exponents in front of the logarithms
$\left(3 - 2 x\right) \cdot \log 4 = - x \log 5$
3. Distribute
$3 \log 4 - 2 x \log 4 = - x \log 5$
4. Rearrange the terms so that each "x" appears on the same side of the equation
$3 \log 4 = 2 x \log 4 - x \log 5$
5. Factor out x from the right side of the equation
$3 \log 4 = x \left(2 \log 4 - \log 5\right)$
6. Isolate (solve for) the variable x by dividing both sides of the equation by $\left(2 \log 4 - \log 5\right)$.
$\frac{3 \log 4}{2 \log 4 - \log 5} = \frac{x \cancel{\left(2 \log 4 - \log 5\right)}}{\cancel{2 \log 4 - \log 5}}$
7. (Optional) Plug into calculator to approximate a value for x
x=(3log4)/(2log4-log5)≈3.5755