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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

Enter the expression you want to simplify into the editor. The simplification calculator allows you to take a simple or complex expression and simplify and reduce the expression to it's simplest form. The calculator works for both numbers and expressions containing variables. Click the blue arrow to submit and see the result!

Here we will show you how to solve 2x = 27. In other words, we will solve for x in order to find the value of x in this equation: 2x = 27 The goal is to get x by itself on the left side. First, we divide both sides by 2 like this:

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x^2: x^{\msquare} \log_{\msquare} \sqrt{\square} \nthroot[\msquare]{\square} \le \ge \frac{\msquare}{\msquare} \cdot \div: x^{\circ} \pi \left(\square\right)^{'} \frac{d}{dx} \frac{\partial}{\partial x} \int \int_{\msquare}^{\msquare} \lim \sum \infty \theta (f\:\circ\:g) f(x)

How do you solve \displaystyle{2}^{{x}}={27} ? https://socratic.org/questions/how-do-you-solve-2-x-27 The answer is \displaystyle{{\log}_{{2}}{\left({27}\right)}} Explanation: To solve this problem we need the inverse operation of exponents: logarithms.