6^x-5=36 equation
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The solution
Detail solution
Given the equation:
$$6^{x} - 5 = 36$$
or
$$\left(6^{x} - 5\right) - 36 = 0$$
or
$$6^{x} = 41$$
or
$$6^{x} = 41$$
- this is the simplest exponential equation
Do replacement
$$v = 6^{x}$$
we get
$$v - 41 = 0$$
or
$$v - 41 = 0$$
Move free summands (without v)
from left part to right part, we given:
$$v = 41$$
We get the answer: v = 41
do backward replacement
$$6^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(6 \right)}}$$
The final answer
$$x_{1} = \frac{\log{\left(41 \right)}}{\log{\left(6 \right)}} = \frac{\log{\left(41 \right)}}{\log{\left(6 \right)}}$$
Sum and product of roots
[src]
$$\frac{\log{\left(41 \right)}}{\log{\left(6 \right)}}$$
$$\frac{\log{\left(41 \right)}}{\log{\left(6 \right)}}$$
$$\frac{\log{\left(41 \right)}}{\log{\left(6 \right)}}$$
$$\frac{\log{\left(41 \right)}}{\log{\left(6 \right)}}$$
log(41)
x1 = -------
log(6)
$$x_{1} = \frac{\log{\left(41 \right)}}{\log{\left(6 \right)}}$$