4^x=8 equation
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The solution
Detail solution
Given the equation:
$$4^{x} = 8$$
or
$$4^{x} - 8 = 0$$
or
$$4^{x} = 8$$
or
$$4^{x} = 8$$
- this is the simplest exponential equation
Do replacement
$$v = 4^{x}$$
we get
$$v - 8 = 0$$
or
$$v - 8 = 0$$
Move free summands (without v)
from left part to right part, we given:
$$v = 8$$
We get the answer: v = 8
do backward replacement
$$4^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(4 \right)}}$$
The final answer
$$x_{1} = \frac{\log{\left(8 \right)}}{\log{\left(4 \right)}} = \frac{3}{2}$$
Sum and product of roots
[src]
3 log(8) pi*I
- + -------- + ------
2 2*log(2) log(2)
$$\frac{3}{2} + \left(\frac{\log{\left(8 \right)}}{2 \log{\left(2 \right)}} + \frac{i \pi}{\log{\left(2 \right)}}\right)$$
3 log(8) pi*I
- + -------- + ------
2 2*log(2) log(2)
$$\frac{\log{\left(8 \right)}}{2 \log{\left(2 \right)}} + \frac{3}{2} + \frac{i \pi}{\log{\left(2 \right)}}$$
/ log(8) pi*I \
3*|-------- + ------|
\2*log(2) log(2)/
---------------------
2
$$\frac{3 \left(\frac{\log{\left(8 \right)}}{2 \log{\left(2 \right)}} + \frac{i \pi}{\log{\left(2 \right)}}\right)}{2}$$
9 3*pi*I
- + --------
4 2*log(2)
$$\frac{9}{4} + \frac{3 i \pi}{2 \log{\left(2 \right)}}$$
$$x_{1} = \frac{3}{2}$$
log(8) pi*I
x2 = -------- + ------
2*log(2) log(2)
$$x_{2} = \frac{\log{\left(8 \right)}}{2 \log{\left(2 \right)}} + \frac{i \pi}{\log{\left(2 \right)}}$$
x2 = log(8)/(2*log(2)) + i*pi/log(2)
x2 = 1.5 + 4.53236014182719*i
x2 = 1.5 + 4.53236014182719*i