The concentration CCC of a chemical in a solution (in mol/L) is modeled by C(n)=3ln(n+4)−2C(n) = 3\ln(n + 4) - 2C(n)=3ln(n+4)−2 for n>−4n > -4n>−4, where nnn is the volume of reagent added in milliliters.
Find an expression for the inverse function C−1(x)C^{-1}(x)C−1(x).
Tick (✓\checkmark✓) one box.
□C−1(x)=ex−23+4\square \quad C^{-1}(x) = e^{\frac{x-2}{3}} + 4□C−1(x)=e3x−2+4
□C−1(x)=ex+23−4\square \quad C^{-1}(x) = e^{\frac{x+2}{3}} - 4□C−1(x)=e3x+2−4
□C−1(x)=13ln(x+4)−2\square \quad C^{-1}(x) = \frac{1}{3\ln(x+4)-2}□C−1(x)=3ln(x+4)−21
□C−1(x)=e3x+2−4\square \quad C^{-1}(x) = e^{3x+2} - 4□C−1(x)=e3x+2−4
216 exam-style questions on WJEC A Level Maths 3.2 Algebra and Functions (A-level only), covering 3.2.1 Algebra and Functions (A-level only), 3.2.2 Algebra and Functions (A-level only), 3.2.3 Algebra and Functions (A-level only), 3.2.4 Algebra and Functions (A-level only), 3.2.5 Algebra and Functions (A-level only), 3.2.6 Algebra and Functions (A-level only), and 3.2 Algebra and Functions (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.