SOLUTION: Solve. Round any approximate solution to the fourth decimal place. 5^x+7=50-3(5)^x Please show me step by step to get the answer. Thanks

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Solve. Round any approximate solution to the fourth decimal place. 5^x+7=50-3(5)^x Please show me step by step to get the answer. Thanks      Log On


   

Question 475932: Solve. Round any approximate solution to the fourth decimal place.
5^x+7=50-3(5)^x
Please show me step by step to get the answer. Thanks
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Solve. Round any approximate solution to the fourth decimal place.
5^x+7=50-3(5^x)
Please show me step by step to get the answer. Thanks
**
5^x+7=50-3(5^x)
4(5^x)=43
5^x=43/4
take log of both sides
x log 5=log 43-log 4
x=(log 43-log 4)/log 5
x=1.4756