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

##### 1 Answer

Jul 26, 2017

#### Explanation:

- Take the common logarithm of both sides

#log4^(3-2x) = log5^-x# - Move the exponents in front of the logarithms

#(3-2x)*log4 = -xlog5# - Distribute

#3log4-2xlog4=-xlog5# - Rearrange the terms so that each "x" appears on the same side of the equation

#3log4=2xlog4-xlog5# - Factor out x from the right side of the equation

#3log4=x(2log4-log5)# - Isolate (solve for) the variable x by dividing both sides of the equation by
#(2log4-log5)# .

#(3log4)/(2log4-log5)=(xcancel((2log4-log5)))/(cancel(2log4-log5))# - (Optional) Plug into calculator to approximate a value for x

#x=(3log4)/(2log4-log5)≈3.5755#