SOLUTION: These questions confuse me a bit and this site has helped me before. Thank you, please show me how to solve these problems (for x) . 16) log(x+4)-log6=1 18) log(x-1)^2=log(-5x-1)

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Question 748655: These questions confuse me a bit and this site has helped me before. Thank you, please show me how to solve these problems (for x) .
16) log(x+4)-log6=1
18) log(x-1)^2=log(-5x-1)
please help. thank you again so much.

Found 2 solutions by stanbon, MathTherapy:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
log(x+4)-log6=1
log[(x+4)/6] = 1
(x+4)/6 = 10^1
x+4 = 60
x = 56
----------------------------
18) log(x-1)^2=log(-5x-1)
(x-1)^2 = -5x-1
x^2 -2x + 1 = -5x -1
-----
x^2 +3x = 0
x(x+3) = 0
x = 0 or x = -3
------------------
Cheers,
Stan H.

Answer by MathTherapy(10556)   (Show Source): You can put this solution on YOUR website!

please show me how to solve these problems (for x) .
16) log(x+4)-log6=1
18) log(x-1)^2=log(-5x-1)
please help. thank you again so much.


16) log(x+4)-log6=1









x + 4 = 60 ------- Cross-multiplying

x = 60 - 4, or


18) log(x-1)^2=log(-5x-1)











(x + 1)(x + 2) = 0

, or

You can do the check on both!!

Send comments and “thank-yous” to “D” at MathMadEzy@aol.com

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