4^(x-1)=1 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$4^{x - 1} = 1$$
or
$$4^{x - 1} - 1 = 0$$
or
$$\frac{4^{x}}{4} = 1$$
or
$$4^{x} = 4$$
- this is the simplest exponential equation
Do replacement
$$v = 4^{x}$$
we get
$$v - 4 = 0$$
or
$$v - 4 = 0$$
Move free summands (without v)
from left part to right part, we given:
$$v = 4$$
We get the answer: v = 4
do backward replacement
$$4^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(4 \right)}}$$
The final answer
$$x_{1} = \frac{\log{\left(4 \right)}}{\log{\left(4 \right)}} = 1$$
$$x_{1} = 1$$
pi*I
x2 = 1 + ------
log(2)
$$x_{2} = 1 + \frac{i \pi}{\log{\left(2 \right)}}$$
Sum and product of roots
[src]
pi*I
1 + 1 + ------
log(2)
$$1 + \left(1 + \frac{i \pi}{\log{\left(2 \right)}}\right)$$
$$2 + \frac{i \pi}{\log{\left(2 \right)}}$$
$$1 + \frac{i \pi}{\log{\left(2 \right)}}$$
$$1 + \frac{i \pi}{\log{\left(2 \right)}}$$
x2 = 1.0 + 4.53236014182719*i
x2 = 1.0 + 4.53236014182719*i