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) About Me  (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) About Me  (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?

8%5E%282x+%2B+1%29+=+4%5E%281+-+x%29

Change current bases to same base:
8%5E%282x+%2B+1%29 becomes %282%5E3%29%5E%282x+%2B+1%29, and 4%5E%281+-+x%29 becomes %282%5E2%29%5E%281+-+x%29

We now have: %282%5E3%29%5E%282x+%2B+1%29+=+%282%5E2%29%5E%281+-+x%29 ----- 2%5E%283%282x+%2B+1%29%29+=+2%5E%282%281+-+x%29%29 ---- 2%5E%286x+%2B+3%29+=+2%5E%282+-+2x%29

Having the same base, this means that: 6x + 3 = 2 - 2x

6x + 2x = 2 - 3

8x = - 1

highlight_green%28x+=+%28-+1%2F8%29_or_0.125%29

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