2^x=16^2 equation
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The solution
Detail solution
Given the equation:
$$2^{x} = 16^{2}$$
or
$$2^{x} - 16^{2} = 0$$
or
$$2^{x} = 256$$
or
$$2^{x} = 256$$
- this is the simplest exponential equation
Do replacement
$$v = 2^{x}$$
we get
$$v - 256 = 0$$
or
$$v - 256 = 0$$
Move free summands (without v)
from left part to right part, we given:
$$v = 256$$
We get the answer: v = 256
do backward replacement
$$2^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(2 \right)}}$$
The final answer
$$x_{1} = \frac{\log{\left(256 \right)}}{\log{\left(2 \right)}} = 8$$
Sum and product of roots
[src]
$$\left(8\right)$$
$$8$$
$$\left(8\right)$$
$$8$$