Question

How do you solve this equation: 4^(2x+3)=1

Answer 1

Hi,

Work:

Equation;


`{4}^(2x + 3) = 1`

Write the number in exponential form with a base of 4.


`{4}^(2x + 3) = 4^(0)`

Since the bases are the same, set the exponents equal.


`2x + 3 = 0`

Move constant +3 to the right side and change its sign.


`2x = - 3`

Divide both sides of equation by 2.


`x = - (3)/(2) \: \: \: \: \: \: \: result \: (fraction) \\ \\ x = - 1.5 \: \: \: \: \: \: \: \: result \: (decimal)`

Hope this helps.

Answer 2

`4^(2x+3)=1\ \ \ \ \ |\log_4\\\\\log_44^(2x+3)=\log_41\\\\\boxed{use\ \log_a1=x\to x=0\ and\ \log_ab^n=n\log_ab}\\\\(2x+3)\log_44=0\\\\\boxed{use\ \log_aa=1}\\\\2x+3=0\ \ \ \ |-3\\\\2x=-3\ \ \ \ |:2\\\\\boxed{x=-1.5}`