Solving (x-6)(x+7)/(x-2) ≥ 0 & log(base 3)x + log(base 3)(x-6) = 3

[Help_Me_Please]

This should be rather easy for the rest of you .. but somehow I can't remember what to do.

solve
[ (x-6) (x+7) ] / (x-2) is greater than or equal to 0

and

solve

log(base three)x + log(base three)(x-6) = 3

ty in advance =)

 

[courtrigrad]

what have you done so far?

\frac{(x-6)(x+7)}{x-2} \geq 0
what are the critical numbers( make numerator 0 or denominator 0)

Then form your test intervals and you get the answer.

for \log_3x + \log_3(x-6) = 3 use log rules

hint: \log_bxy = log_bx + log_by

 

[wais]

To solve (x-6)(x+7)/(x-2) ≥ 0, we can start by finding the critical points, which are the values of x that make the inequality equal to 0 or undefined. In this case, the critical points are x=2, x=6, and x=-7.

Next, we can create a number line and plot these critical points on it. Then, we can test a value in each of the intervals created by these points to see if it satisfies the inequality. For example, testing x=0 in the interval (-∞, -7) would give us a negative result, so this interval would not satisfy the inequality. Similarly, testing x=4 in the interval (-7, 2) would give us a positive result, so this interval would satisfy the inequality.

Therefore, the solution to the inequality is (-∞, -7] ∪ [2, 6] ∪ (6, ∞). This means that all values of x in these intervals will make the inequality true.

To solve log(base 3)x + log(base 3)(x-6) = 3, we can use the logarithmic properties to simplify the equation. First, we can combine the two logarithms using the product rule, which states that log(a) + log(b) = log(ab). In this case, we get log(base 3)(x(x-6)) = 3.

Next, we can rewrite the equation in exponential form, which states that log(base a)b = c is equivalent to a^c = b. In this case, we get 3^3 = x(x-6).

Solving for x, we get x=9 or x=-3. However, we have to check these solutions in the original equation to make sure they are valid. Plugging in x=9 gives us log(base 3)9 + log(base 3)(9-6) = 3, which is true. However, plugging in x=-3 gives us log(base 3)(-3) + log(base 3)(-3-6) = 3, which is not defined.

Therefore, the only valid solution is x=9.