The figures below show the graphs of the exponential functio

The figures below show the graphs of the exponential functions f(x) and g(x), and the linear function, h(x). The function f(x) has y-intercept 0.75 and goes through the point (1, 6). The function g(x) has y-intercept 2 and goes through the point (2, 2/9). The function h(x) has y-intercept 3 and goes through the point (a, a + 3). (a) Find a formula for f(x) = 0.75*8^x (b) Find a formula for g(x) = 2*0.33333^x (c) Find a formula for h(x) = x+3 (d) Find the exact value(s) of x such that f(x) = g(x). If there is more than one solution, enter your answers as a comma separated list. X =

Solution

(a). Let f(x) = m*n^x ,where m, n are arbitrary real numbers. The y-intercept for f(x) is 0.75 i.e. f(x) = 0.75 when x = 0. On substituting these values of x and f(x) in the equation for f(x), we get 0.75 = m*n⁰ = m ( as n⁰ =1). Then f(x) = 0.75*n^x. Further, the graph of f(x) passes through the point (1,6) so that 6 = 0.75*n¹ or, n = 6/0.75 = 8. Hence f(x) = 0.75*8^x.

(b). Let g(x) = p*q^x, where p, q are arbitrary real numbers. The y-intercept for h(x) is 2 i.e. g(x) = 2 when x = 0. On substituting these values of x and g(x) in the equation for g(x), we get 2 = p*q⁰ = p ( as q⁰ =1). Then g(x) = 2*q^x. Further, the graph of g(x) passes through the point (2,2/9) so that 2/9= 2*q² or, q² = (2/9)*1/2 = 1/9 so that q = +1/3 = 0.033333( aproximately)( If q is negative, then the y-intercept of g(x) will be negative,which is not the case here). Thus, g(x)= 2*(1/3)^x = 2*(0.033333)^x.

(c). The graph of h(x) is a line. Let h(x) = mx+c, where mm is its slope and c is its y-intercept. Here, c = 3 so that h(x) = mx+3. Further, the graph of h(x) passes through the point (a,a+3). On substituting these values of x and h(x) in the equation for h(x), we get a+3 = am +3 or, am = a+3-3 = a so that m = a/a =1. Then h(x) = x+3.

(d). If f(x) = g(x), then 0.75*8^x= 2*(1/3)^x or, 8^x * 3^x = 2/0.75 or, 24^x = 2/0.75.Now, on taking logarithm of both the sides, we have log 24^x = log(2/0.75) or, x log 24 = log 2 –log 0.75 so that x = (log2–log0.75)/log24=[0.301029995–(-0.124938736)]/1.380211242 = 0.425968731/1.380211242 = 0.30862575 = 0.308626 ( on rounding off to 6 decimal places).

Note: log mⁿ = n log m and log (p/q) = log p – logq.