SOLUTION: How do you solve: {{{ 4^(x-4)=0.5^(5-2x) }}}.

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: How do you solve: {{{ 4^(x-4)=0.5^(5-2x) }}}.      Log On


   

Question 249342: How do you solve: +4%5E%28x-4%29=0.5%5E%285-2x%29+.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
+4%5E%28x-4%29=0.5%5E%285-2x%29+ Start with the given equation.


+%282%5E2%29%5E%28x-4%29=%281%2F2%29%5E%285-2x%29+ Rewrite 4 as 2%5E2. Rewrite 0.5 as 1%2F2


+%282%5E2%29%5E%28x-4%29=%282%5E%28-1%29%29%5E%285-2x%29+ Rewrite 1%2F2 as 2%5E%28-1%29


+2%5E%282%28x-4%29%29+=2%5E%28-%285-2x%29%29+ Multiply the exponents.


2%28x-4%29=-%285-2x%29 Since the bases are equal, the exponents are equal.


2x-8=-5%2B2x Distribute.


2x=-5%2B2x%2B8 Add 8 to both sides.


2x-2x=-5%2B8 Subtract 2x from both sides.


0x=-5%2B8 Combine like terms on the left side.


0x=3 Combine like terms on the right side.


0=3 Simplify.


Since this equation is never true for any x value, this means that there are no solutions.


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Answer:

So there are no solutions to the equation +4%5E%28x-4%29=0.5%5E%285-2x%29+