Exponentials and Logarithms

a.

Using $y = 2^x$ as a substitution, show that $4^x - 2^{x+2} - 5 = 0$ can be written as $y^2 - 4y - 5 = 0$ .

[2]

b.

Hence, show that the equation $4^x - 2^{x+2} - 5 = 0$ has $x = \log_2 5$ as its only solution.

[4]

a.

Using $y = 2^x$ as a substitution, show that $4^x - 2^{x+2} - 5 = 0$ can be written as $y^2 - 4y - 5 = 0$ .

[2]

b.

Hence, show that the equation $4^x - 2^{x+2} - 5 = 0$ has $x = \log_2 5$ as its only solution.

[4]

Exponentials and Logarithms Questions