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You have found 3^{x-2}=\frac{12}{9}=\frac{4}{3} . So replace the \frac{4}{3} in your expression: \left(\sqrt{\frac43}\right)^{1/(x-2)}=\left(\sqrt{3^{x-2}}\right)^{1/(x-2)}=\cdots

f(x)=x^3 ; prove\:\tan^2(x)-\sin^2(x)=\tan^2(x)\sin^2(x) \frac{d}{dx}(\frac{3x+9}{2-x}) (\sin^2(\theta))' \sin(120) \lim _{x\to 0}(x\ln (x)) \int e^x\cos (x)dx \int_{0}^{\pi}\sin(x)dx

Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor.

It's easy to prove that 3x ≥ 9x for all real x ≤ 0 and x ≥ 3 with equality at x = 3. Furthermore, in the interval (0, 3) there is exactly one α such that 3α = 9α. Put all this together and it's not hard to …

A Nice Exponential Equation 3^x = X^9 How to Solve Math Olympiad Question 3^x = X^9 Exponential Equation? What is the value of x? Find the value of x? Solvin...

What is the answer to 3^x=-9 ?

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