Question

How do I get the bottom x away? 3=x^2-6x+8/x

Answer 1

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Solve the logarithmic equation

log₃(x – 4) + log₃(x – 2) – log₃ x = 1


Condition:  x > 4,  because logarithms are only defined for positive numbers.

log₃(x – 4) + log₃(x – 2) – log₃ x = 1

log₃(x – 4) + log₃(x – 2) – log₃ x = log₃ 3


Applying log properties, you can rewrite that equation as

$\mathsf{log_3\!\left[\dfrac{(x-4)(x-2)}{x} \right ]=log_3\,3}$


Since log is an one-to-one function, you can "cancel out" those logs at both sides, so you have

$\mathsf{\dfrac{(x-4)(x-2)}{x}=3}$


Multiply both sides by  x  to simplify that denominator:

$\mathsf{\diagup\!\!\!\! x\cdot \dfrac{(x-4)(x-2)}{\diagup\!\!\!\! x}=x\cdot 3}\\\\\\ \mathsf{(x-4)(x-2)=3x}$


Multiply out those brackets, by applying the distributive property:

$\mathsf{(x-4)\cdot x-(x-4)\cdot 2=3x}\\\\ \mathsf{x^2-4x-2x+8=3x}$


Subtract  3x  from both sides, and then combine like terms together:

$\mathsf{x^2-4x-2x+8-3x=3x-3x}\\\\ \mathsf{x^2-4x-2x-3x+8=0}\\\\ \mathsf{x^2-9x+8=0}$


Now you have a quadratic equation, where the coefficients are

a = 1,  b = – 9,  c = 8

Solve it using the quadratic formula:


Finding the discriminant  Δ:

Δ = b² – 4ac

Δ = (– 9)² – 4 · 1 · 8

Δ = 81 – 32

Δ = 49

Δ = 7²


Then,

$\mathsf{x=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\ \mathsf{x=\dfrac{-(-9)\pm\sqrt{7^2}}{2\cdot 1}}\\\\\\ \mathsf{x=\dfrac{9\pm 7}{2}}$

$\begin{array}{rcl} \mathsf{x=\dfrac{9-7}{2}}&~\textsf{ or }~&\mathsf{x=\dfrac{9+7}{2}}\\\\ \mathsf{x=\dfrac{2}{2}}&~\textsf{ or }~&\mathsf{x=\dfrac{16}{2}}\\\\ \mathsf{x=1}&~\textsf{ or }~&\mathsf{x=8} \end{array}$


You can discard  x = 1  as a solution, because those initial logarithms are not defined for this value of x (remember that  x  must be greater than 4).

So the only solution is  x = 8.


Solution set:  S = {8}.


I hope this helps. =)


Tags:  logarithmic logarithm log equation condition quadratic formula discriminant solve algebra