Question

Solve for xLog base 6 x + log base 6 (x-1)=1

Answer 1

Hello there. To solve this question, we have to remember some properties about logarithmic equations.

Given the following equation:

$\log_6(x)+\log_6(x-1)=1$

We want to solve it for x.

First, apply the property:

$\log_a(b)+\log_a(c)=\log_a(b\cdot c)$

Hence we'll get

$\log_6(x\cdot(x-1))=1$

Now, apply the following property:

$\log_a(b)=c\Rightarrow b=a^c$

Therefore we get

$x\cdot(x-1)=6^1=6$

Apply the FOIL:

$x^2-x=6$

Subtract 6 on both sides of the equation

$x^2-x-6=0$

To solve the quadratic equation, use the formula:

For ax² + bx + c = 0 and a not equal to zero,

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

Hence we'll have a = 1, b = -1 and c = 6, so

$x=\dfrac{-(-1)\pm\sqrt{(-1)^2-4\cdot1\cdot(-6)}}{2\cdot1}=\dfrac{1\pm\sqrt{1+24}}{2}$

Adding the values in the radical, we get

$x=\dfrac{1\pm\sqrt{25}}{2}$

Knowing 25 = 5², we have

$x=\dfrac{1\pm5}{2}$

Separating the solutions, we get

$\begin{gathered} x_1=\dfrac{1+5}{2}=\dfrac{6}{2}=3 \\ \\ x_2=\dfrac{1-5}{2}=\dfrac{-4}{2}=-2 \end{gathered}$

But the argument of a logarithm has to be greater than zero, so

$x=3$

Is the only solution to this logarithmic equation.