![\text{Given that,}\\\\f(x) = ab^x, f(0) = -5 ~ \text{And}~ f(3) = - \dfrac 5{27}\\\\\\\text{Now,}\\\\\\f(0) = -5 \\\\\implies ab^0 = -5\\\\\implies a =-5\\\\\\f(3) = -\dfrac{5}{27}\\\\\implies ab^3 = -\dfrac{5}{27}\\\\\implies -5b^3 = -\dfrac{5}{27}\\\\\implies b^3 = \dfrac 1{27}\\\\\\\implies b = \sqrt[3]{\dfrac{1}{27}} = \dfrac 13\\\\\\\text{Hence , } f(x) = ab^x = -5\left(\dfrac 13 \right)^x](/tpl/images/2597/0686/e9f57.png)
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