Question

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Answer 1

Given:

Exponential functions

$2^x$, $3\cdot 2^x$ and $2^{3x}$

To find:

The value of a, b, c, d, e and f.

Solution:

Substitute x = 0 in $2^x$.

$a=2^0$

a = 1    (∵ $a^0=1$)

Substitute x = 0 in $3\cdot 2^x$.

$b=3\cdot 2^0$

$b = 3\cdot 1$  (∵ $a^0=1$)

b = 3

Substitute x = 0 in $2^{3x}$.

$c=2^{3\cdot 0}$

$c= 2^{0}$  (∵ $a^0=1$)

c = 1

Substitute x = 1 in $2^x$.

$d=2^1$

d = 2

Substitute x = 1 in $3\cdot 2^x$.

$e=3\cdot 2^1$

$e = 3\cdot 2$

e = 6

Substitute x = 1 in $2^{3x}$.

$f=2^{3\cdot 1}$

$f= 2^{3}$

f = 8

Hence a = 1, b = 3, c = 1, d = 2, e = 6 and f = 8.