answer:The equation is equivalent to: log_7 (x+14)+log_7 (24 x-5)=log_7 (-22 x-25) log_7 [(x+14)(24 x-5)]=log_7 (-22 x-25) (x+14)(24 x-5)=-22 x-25 24 x^2+336 x-35 x-70=-22 x-25 24 x^2+301 x-45=0 (24 x-15)(x+3)=0 x=frac{15}{24} quad text{or} quad x=-3 Since the logarithm function is defined only for positive numbers, we must discard the solution x=-3. Therefore, the only real solution is x=frac{15}{24}.

answer:Hint: Rewrite the expression as lim_{xto 0^-} frac{e^{1/x}}{x} = lim_{xto 0^-} frac{frac{1}{x}}{frac{1}{e^{1/x}}} Now, apply L'Hopital's rule: lim_{xto 0^-} frac{e^{1/x}}{x} = lim_{xto 0^-} frac{frac{d}{dx}(frac{1}{x})}{frac{d}{dx}(frac{1}{e^{1/x}})} = lim_{xto 0^-} frac{-frac{1}{x^2}}{-frac{1}{x^2}e^{1/x}} = lim_{xto 0^-} frac{e^{1/x}}{x^2} Now, we can evaluate this limit directly: lim_{xto 0^-} frac{e^{1/x}}{x^2} = frac{e^{infty}}{0^2} = frac{infty}{infty} Since this is still an indeterminate form, we can apply L'Hopital's rule again: lim_{xto 0^-} frac{e^{1/x}}{x^2} = lim_{xto 0^-} frac{frac{d}{dx}(e^{1/x})}{frac{d}{dx}(x^2)} = lim_{xto 0^-} frac{frac{1}{x}e^{1/x}}{2x} = lim_{xto 0^-} frac{e^{1/x}}{2x^2} Now, we can evaluate this limit directly: lim_{xto 0^-} frac{e^{1/x}}{2x^2} = frac{e^{infty}}{2(0)^2} = frac{infty}{infty} Since this is still an indeterminate form, we can apply L'Hopital's rule a third time: lim_{xto 0^-} frac{e^{1/x}}{2x^2} = lim_{xto 0^-} frac{frac{d}{dx}(e^{1/x})}{frac{d}{dx}(2x^2)} = lim_{xto 0^-} frac{frac{1}{x}e^{1/x}}{4x} = lim_{xto 0^-} frac{e^{1/x}}{4x^2} Now, we can finally evaluate this limit: lim_{xto 0^-} frac{e^{1/x}}{4x^2} = frac{e^{infty}}{4(0)^2} = frac{infty}{infty} = 0 Therefore, lim_{xto 0^-} frac{e^{1/x}}{x} = 0

answer:The reaction depicted in the image is an example of an SN2 nucleophilic substitution reaction. In this mechanism, the nucleophile (OH-) attacks the substrate (CH3Br) from the backside, opposite to the leaving group (Br-). This concerted process results in the inversion of configuration at the carbon center, forming CH3OH and releasing Br- as the leaving group.

answer:We have that: 2^{4.5} = 2^{3}cdot 2^{1.5} = 8cdot sqrt{2^{3}} > 8cdot sqrt{2^{2}} = 8cdot 2 = 16. On the other hand, 9^{1.5} = 9^{0.5}cdot 9 = 3cdot 9 = 27. Since 16 < 27, we conclude that 2^{4.5} > 9^{1.5}.