Combine Logbase3 X^2 3 Logbase3 X-2 1/3 Logbase3 X 1 Into A Single Logarithm.

How can I solve for x if log base 9 of x + log base 3 of x = 6?

Thanks for the A2A!Note that [math]\log_9{x}=\frac{1}{2}\log_3{x}[/math]. So:[math]\frac{1}{2}\log_3{x}+\log_3{x}=6\tag*{}[/math]We can simplify the LHS as:[math]\frac{3}{2}\log_3{x}=6\tag*{}[/math]Multiplying both sides by [math]\frac{2}{3}[/math]:[math]\log_3{x}=4\tag*{}[/math]Finally, raising [math]3[/math] to both sides:[math]x=3^4=9^2=81\tag*{}[/math]You can verify this by plugging it in.

How do you find the derivative of log base a of x?

You can express [math]\log_a(x)[/math] as [math]\log_a(x)= \ln(x)/\ln(a)[/math]. (To show this use the identity [math]x = a^{log_a(x)} = e^{\ln(x)}[/math] and take the natural logarithm of the last equation.) So the derivative is[math][\log_a(x)]^\prime = \frac{1}{x\ln(a)}[/math]

What is the inverse of y = log 3 x?

First of all, we know that function f represented by the given logarithmic equation, y = f(x) = log 3x, has an inverse, f ̄¹ (read “f inverse” or “the inverse of f”¹), since logarithmic function f is one-to-one, i.e., no two ordered pairs (x, y) of function f have the same second component.In finding the inverse of the given logarithmic function f, we must first interchange the x and y variables in the given equation defining function f as follows:(1.) y = log 3x, which is equivalent to 10ʸ = 3x, becomes:(2.) x = log 3y, since the ordered pairs between the two functions, f and f ̄¹, are reversed, i.e., if (x₁, y₁) is an ordered pair of f, then (y₁, x₁) is an ordered pair of f ̄¹. ² “The domain and range of f ̄¹ are the range and domain, respectively, of f.” ³Now, the next step is to solve equation (2.) for y in terms of x, as follows:The logarithmic equation x = log 3y is equivalent to its corresponding exponential form:(3.) 10ˣ = 10^(log 3y) , which becomes:(4.) 10ˣ = 3y(I chose the base to be 10 because my experience and background has been such that if no base is specified for a given logarithm, then the base is understood to be 10 (common logarithms); however, any valid base could have been used, but needs to be specified, for example, if the irrational number “e” is the base, then use either logₑ or ln)(5.) 3y = 10ˣ (since equality is symmetric, i.e., if a = b, then b = a)Now, dividing both sides by 3, we get:(6.) 3y/3 = 10ˣ/3(7.) (3/3)y = 10ˣ/3 (8.) (1)y = 10ˣ/3(9.) y = f ̄¹(x) = 10ˣ/3 is the inverse of y = f(x) = log 3x.VERIFICATION:Two functions are inverses of each other if they undo each other, i.e., if only the original input variable x is remaining; This undoing is demonstrated as follows by the operation of composition in which functions f and f ̄¹ are combined together to produce a third function, i.e., f ⃘ f ̄¹ or f ̄¹ ⃘ f:⁴(f ⃘ f ̄¹)(x) = f(f ̄¹(x))  = f(10ˣ/3)  = log [3(10ˣ/3)] = log [(1)(10ˣ)] = log 10ˣ = x for all x in the domain of f ̄¹, and …(f ̄¹ ⃘ f )(x) = f ̄¹(f(x))  = f ̄¹(log 3x) = [10^(log 3x)]/3 = 3x/3 = (3/3)x = (1)x = x for all x in the domain of f .Therefore, y = f(x) = log 3x and y = f ̄¹(x) = 10ˣ/3 are indeed inverses of each other.¹Jerome E. Kaufmann, “Algebra with Trigonometry for College  Students, third edition, PWS-KENT Publishing Company,  Boston, 1992, p. 404. ² Ibid.³ Ibid.⁴ Ibid., p. 405.

How do I find the domain of f(x) = √ [((x-1) (x+2)) / ((x-3) (x-4))]?

Assuming you are looking for answers such that [math]x \in R[/math], we can use a simple step by step analysis:Anything inside a square root (radical) cannot be negative. We are going to keep this in mind.Through point 1, we can infer that the only part of importance in this question is what is inside the radical sign. So let us continue our analysis to [math]\frac{(x-1)(x+2)}{(x-3)(x-4)}[/math]. Since its a purely rational expression, I'm going to use a simple zero/pole analysis. A zero is a value of the independent variable, that makes the expression zero. Similarly, a pole makes the expression undefined (tend to infinity: [math]\rightarrow \infty [/math]). So our expression yields: Zeroes: [math]x \in \left \{ 1, -2 \right \}[/math]Poles: [math]x \notin \left \{ 3, 4 \right \}[/math]4.  It is obvious that since we need a real output, our independent variable 'x' can never take values {3, 4}. Also our domain space is now divided into five parts: [math]-\infty < x \leq -2[/math][math]-2 \leq x \leq 1[/math][math]1 \leq x < 3[/math][math]3 < x < 4[/math][math]4 < x < \infty[/math]5. Now we simply analyse these domain spaces, one-by-one. I'll do one for you (sweat out the rest yourself): [math]1 \leq x < 3[/math]. Now in this case, the terms (x-1) and (x+2) are positive while (x-3) and (x-4) are (obviously) negative terms. This makes the complete expression of Step 2, positive. Since this result confirms with all of our steps above, we can safely agree that this space belongs in our domain. Similarly, we can see that the interval [math]-2 \leq x \leq 1[/math] cannot belong to our domain.6. The end result is that our domain is the union of all valid spaces. In this particular case, it comes out to be [math](-\infty , -2] \cup [1, 3) \cup (4, \infty)[/math]. NOTE: A generalised shortcut to the above approach is the wavy-curve method. Do note that the wavy curve method must be applied very carefully and is valid only for rational expressions.PS: If you liked my answer, do upvote. If you find a blooper, comment or suggest an edit.