SOLUTION: 4(2^x)+31(2^x)-8=0

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: 4(2^x)+31(2^x)-8=0      Log On


   



Question 256102: 4(2^x)+31(2^x)-8=0
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
4%282%5Ex%29%2B31%282%5Ex%29-8=0 Start with the given equation.


Let z=2%5Ex


4z%2B31z-8=0 Replace each 2%5Ex term with 'z'


35z-8=0 Combine like terms.


35z=8 Add 8 to both sides.


z=8%2F35 Divide both sides by 35.


2%5Ex=8%2F35 Plug in z=2%5Ex


Recall that b%5Ex=y can be rewritten as log%28b%2C%28y%29%29=x


log%282%2C%288%2F35%29%29=x Use the property shown above to rewrite the equation.


x=log%282%2C%288%2F35%29%29 Rearrange the equation.


So the exact answer is x=log%282%2C%288%2F35%29%29 which approximates to x=-2.12928