Question

Exponential Equations Not Requiring Logarithms


solve for p

(2^-3p) . (2^2p) = (2^2p)

Answer 1

The way I think to solve this is this way:

Since the base is the same.

-3p + 2p = 2p

-p= 2p How is this possible? What should I write for the answer:?

Answer 2

yes\[%(2^{-3p}) \cdot (2^{2p}) = (2^{2p})\\
%2^{-3p+2p}=2^{2p}\\
%2^{-p}=2^{2p}\\
-p =2p
\]
solve for p

Answer 3

2p +p = 0

3p = 0

p = 0/3 = 0

Answer 4

yeah, and you can check it by substituting p=0 into eh original equation

1 * 1 = 1

Answer 5

Right :)

Answer 6

Thank you for clarifying it.