Given:
Exponential functions
[tex]2^x[/tex], [tex]3\cdot 2^x[/tex] and [tex]2^{3x}[/tex]
To find:
The value of a, b, c, d, e and f.
Solution:
Substitute x = 0 in [tex]2^x[/tex].
[tex]a=2^0[/tex]
a = 1 (∵ [tex]a^0=1[/tex])
Substitute x = 0 in [tex]3\cdot 2^x[/tex].
[tex]b=3\cdot 2^0[/tex]
[tex]b = 3\cdot 1[/tex] (∵ [tex]a^0=1[/tex])
b = 3
Substitute x = 0 in [tex]2^{3x}[/tex].
[tex]c=2^{3\cdot 0}[/tex]
[tex]c= 2^{0}[/tex] (∵ [tex]a^0=1[/tex])
c = 1
Substitute x = 1 in [tex]2^x[/tex].
[tex]d=2^1[/tex]
d = 2
Substitute x = 1 in [tex]3\cdot 2^x[/tex].
[tex]e=3\cdot 2^1[/tex]
[tex]e = 3\cdot 2[/tex]
e = 6
Substitute x = 1 in [tex]2^{3x}[/tex].
[tex]f=2^{3\cdot 1}[/tex]
[tex]f= 2^{3}[/tex]
f = 8
Hence a = 1, b = 3, c = 1, d = 2, e = 6 and f = 8.
