SOLUTION: 8^(2x+1)= 4^(1-x)
In the real numbers, what is the solution of the equation above?
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Question 618295: 8^(2x+1)= 4^(1-x)
In the real numbers, what is the solution of the equation above?
Found 2 solutions by lwsshak3, MathTherapy:
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
8^(2x+1)= 4^(1-x)
In the real numbers, what is the solution of the equation above?
using logarithms:
(2x+1)log8=(1-x)log4
2xlog8+log8=log4-xlog4
2xlog8+xlog4=log4-log8
x(2log8+log4)=(log4-log8)
x=(log4-log8)/(2log8+log4)
using calculator
x=-0.125
Answer by MathTherapy(10555) (Show Source): You can put this solution on YOUR website!
8^(2x+1)= 4^(1-x)
In the real numbers, what is the solution of the equation above?
Change current bases to same base:
becomes , and becomes
We now have: ----- ----
Having the same base, this means that: 6x + 3 = 2 - 2x
6x + 2x = 2 - 3
8x = - 1
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