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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
[tex]\mathsf{log_3\!\left[\dfrac{(x-4)(x-2)}{x} \right ]=log_3\,3}[/tex]
Since log is an one-to-one function, you can "cancel out" those logs at both sides, so you have
[tex]\mathsf{\dfrac{(x-4)(x-2)}{x}=3}[/tex]
Multiply both sides by x to simplify that denominator:
[tex]\mathsf{\diagup\!\!\!\! x\cdot \dfrac{(x-4)(x-2)}{\diagup\!\!\!\! x}=x\cdot 3}\\\\\\ \mathsf{(x-4)(x-2)=3x}[/tex]
Multiply out those brackets, by applying the distributive property:
[tex]\mathsf{(x-4)\cdot x-(x-4)\cdot 2=3x}\\\\ \mathsf{x^2-4x-2x+8=3x}[/tex]
Subtract 3x from both sides, and then combine like terms together:
[tex]\mathsf{x^2-4x-2x+8-3x=3x-3x}\\\\ \mathsf{x^2-4x-2x-3x+8=0}\\\\ \mathsf{x^2-9x+8=0}[/tex]
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,
[tex]\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}}[/tex]
[tex]\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}[/tex]
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
