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

1 Answer
Jul 26, 2017

Answer:

#x=((3log4)/(2log4-log5))≈3.5755#

Explanation:

#4^(3-2x)=5^-x#

  1. Take the common logarithm of both sides
    #log4^(3-2x) = log5^-x#
  2. Move the exponents in front of the logarithms
    #(3-2x)*log4 = -xlog5#
  3. Distribute
    #3log4-2xlog4=-xlog5#
  4. Rearrange the terms so that each "x" appears on the same side of the equation
    #3log4=2xlog4-xlog5#
  5. Factor out x from the right side of the equation
    #3log4=x(2log4-log5)#
  6. 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))#
  7. (Optional) Plug into calculator to approximate a value for x
    #x=(3log4)/(2log4-log5)≈3.5755#