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# logarithms

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What is log base9 (x^2-4x) = log base9 (3x-10)

Guest Feb 27, 2017
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#1
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Solve for x:
(log(x^2 - 4 x))/(log(9)) = (log(3 x - 10))/(log(9))

Subtract (log(3 x - 10))/(log(9)) from both sides:
(log(x^2 - 4 x))/(log(9)) - (log(3 x - 10))/(log(9)) = 0

Bring (log(x^2 - 4 x))/(log(9)) - (log(3 x - 10))/(log(9)) together using the common denominator log(9):
-(log(3 x - 10) - log(x^2 - 4 x))/(log(9)) = 0

Multiply both sides by -log(9):
log(3 x - 10) - log(x^2 - 4 x) = 0

log(3 x - 10) - log(x^2 - 4 x) = log(3 x - 10) + log(1/(x^2 - 4 x)) = log((3 x - 10)/(x^2 - 4 x)):
log((3 x - 10)/(x^2 - 4 x)) = 0

Cancel logarithms by taking exp of both sides:
(3 x - 10)/(x^2 - 4 x) = 1

Multiply both sides by x^2 - 4 x:
3 x - 10 = x^2 - 4 x

Subtract x^2 - 4 x from both sides:
-x^2 + 7 x - 10 = 0

The left hand side factors into a product with three terms:
-(x - 5) (x - 2) = 0

Multiply both sides by -1:
(x - 5) (x - 2) = 0

Split into two equations:
x - 5 = 0 or x - 2 = 0

x = 5 or x - 2 = 0

Answer:      x = 5            or              x = 2(assuming a complex-valued log)

Guest Feb 27, 2017
#2
+79689
0

Since the log bases are the same, we can forget the log part and solve this :

x^2 - 4x   = 3x - 10       subtract 3x from both sides....add 10 to both sides

x^2 - 4x - 3x + 10  = 0

x^2 - 7x + 10  = 0     factor

(x - 5) (x - 2)  = 0

Set both factors to = 0   and x = 5   or x  = 2

However.....we must reject 2 because it makes both logs undefined.....so.....

x = 5  is the solution

CPhill  Feb 27, 2017

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