SOLUTION: solve the equation (2^x).(2^x+1)=10 correct to 3decimal places im getting my answer as 1.161 but it dont tally when i substitute the value for x pliz help

Question 407913: solve the equation (2^x).(2^x+1)=10 correct to 3decimal places
im getting my answer as 1.161 but it dont tally when i substitute the value for x pliz help
Answer by graphmatics(170) (Show Source): You can put this solution on YOUR website!
To solve the equation (2^x).(2^x+1)=10 we first want to try to move the x expression from their exponent positions. The way we will do that is with
log base 2 (check out Log at http://en.wikipedia.org/wiki/Logarithm).
We have that
(2^x).(2^x+1)=10
LogBase2((2^x)(2^x+1)) = LogBase2(10)
LogBase2(2^x) + LogBase2(2^x+1) = LogBase2(10)
xLogBase2(2) + (x+1)LogBase2(2) = LogBase2(10)
x + x+1 = 3.3219
2*x = 2.3219
x = 1.661
You have the answer is 1.161 but I got 1.661

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