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#1
Oct 8 2015 10:34am
log(x+9)=1-log x

Can someone explain on how to solve this problem?
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#2
Oct 8 2015 11:02am
Log(x+9) = log(10) - log(x)

Log(x+9) = log(10/x)

10^both sides

X+9 = 10/x

X = 1
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#3
Oct 8 2015 11:12am
Quote (RzChaos @ Oct 8 2015 01:02pm)
Log(x+9) = log(10) - log(x)

Log(x+9) = log(10/x)

10^both sides

X+9 = 10/x

X = 1


gg
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#4
Oct 8 2015 06:17pm
Quote (RzChaos @ Oct 8 2015 12:02pm)
Log(x+9) = log(10) - log(x)

Log(x+9) = log(10/x)

10^both sides

X+9 = 10/x

X = 1


A bit more explanation in case it wasn't clear

X + 9 = 10/X (multiply both sides by X)
X^2 + 9X = 10 (set equation equal to 0)
X^2 + 9X - 10 = 0 (FOIL)
(X + 10)(X - 1) = 0 --> X = -10 OR X = 1
since you cannot take the log on a negative number, X cannot equal -10 ---> X = 1
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#5
Oct 8 2015 06:48pm
Quote (RzChaos @ Oct 8 2015 10:02am)
Log(x+9) = log(10) - log(x)

Log(x+9) = log(10/x)

10^both sides

X+9 = 10/x

X = 1


Quote (TritonV8 @ Oct 8 2015 05:17pm)
A bit more explanation in case it wasn't clear

X + 9 = 10/X (multiply both sides by X)
X^2 + 9X = 10 (set equation equal to 0)
X^2 + 9X - 10 = 0 (FOIL)
(X + 10)(X - 1) = 0 --> X = -10 OR X = 1
since you cannot take the log on a negative number, X cannot equal -10 ---> X = 1


thanks guys, I appreciate it :)
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