Question

Log6(x-2)+ log6(х-11) =2

Answer 1

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Answer 2

To solve the equation $\log_6(x-2) + \log_6(x-11) = 2$, you can use logarithm properties to simplify it and then solve for $x$. Here's the step-by-step solution:

1. Use the logarithm property $\log_a(x) + \log_a(y) = \log_a(xy)$ to combine the two logarithms on the left side:

$\log_6((x-2)(x-11)) = 2$

2. Remove the logarithm by exponentiating both sides. Since the base is 6, raise 6 to the power of both sides:

$6^{\log_6((x-2)(x-11))} = 6^2$

3. On the left side, the base 6 and logarithm cancel each other out, leaving you with:

$(x-2)(x-11) = 36$

4. Expand the left side of the equation:

$x^2 - 11x - 2x + 22 = 36$

5. Combine like terms:

$x^2 - 13x + 22 = 36$

6. Move 36 to the left side by subtracting it from both sides:

$x^2 - 13x + 22 - 36 = 0$

7. Simplify further:

$x^2 - 13x - 14 = 0$

8. To solve this quadratic equation, you can factor it or use the quadratic formula. In this case, let's factor it:

$(x - 14)(x + 1) = 0$

9. Set each factor equal to zero and solve for $x$:

a) $x - 14 = 0$ $x = 14$

b) $x + 1 = 0$ $x = -1$

So, the solutions to the equation are $x = 14$ and $x = -1$. You should always check if these solutions are valid by plugging them back into the original equation to ensure that the logarithms are defined for those values. In this case, both solutions are valid.