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
-> 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
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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) (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
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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
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